Some Notes on Sequences and Series
prepared for a second course in calculus
Thomas L. Scofield
Department of Mathematics and Statistics
Calvin College
2003
1
The purpose of these notes is to present some of the notions of sequences and series without
going into as great of depth as in your text. These notes may be thought of as self-contained (i.e.,
one does not have to read from other sources to understand what is being said here) but, wherever
appropriate, you will be directed to look at examples from the Thomas Calculus book. You are
also encouraged to read that text to round out your understanding of sequences and series.
1
Sequences
Up to this point in the two semesters of calculus, we have consistently dealt with functions defined
on intervals of numbers or unions of such. Polynomials and exponentials, for instance, are defined
for all real numbers (i.e., on the domain (−∞, ∞)); rational functions accept any input which
does not make the expression in the denominator zero; similar statements may be made about the
trigonometric functions (after writing them in terms of sines and cosines).
For the person thinking about applying mathematics to the world around us, functions are an
idealization of data. A human’s body temperature or pulse may be thought of as a function of time;
the height of a child as well. The air pressure in a tire can be viewed as a function of temperature.
In each of these, it is presumed that, for a continuous interval (domain) of inputs, there is a range of
outputs. Nevertheless, in practice it is rare that one has access to such continuous data. (Note: the
word “continuous” is being used here not to refer to a function itself, but rather in the sense that
there are stretches without break in the domain of that function.) One requires time to measure a
person’s pulse — enough time for one full cycle of atrial and ventricular contractions, at a minimum
— which means that one gets an average heartrate, not a string of heartrates for each moment in
time. The same may be said even if a sensor is used to take the measurements, rather than medical
personnel. Indeed, there is little chance that any sensor in the near future will be able to collect
continuous data of any kind, nor that there will be a computer able to accept and process such
data.
Instead, data generally comes to us in discrete packets or points. We can posit the existence
of a curve that connects or fills in the gaps between points, and set ourselves to finding such a
curve (there is no single answer to that problem), and in this section we will do so whenever it is
convenient. Nevertheless, we might just consider the points themselves — particularly, the values
(y-coordinates, if you prefer thinking of values of a function in that way) at these points.
Often discrete values come to us in some sort of order, and we might label them in the order
a1 = 1st value
a2 = 2nd value
a3 = 3rd value
in which the came:
Though, in all data-collection we have experienced, one
..
.
an
=
..
.
nth value
terminates the process, many processes may be viewed as possibly forever on-going. For instance,
a meteorologist might start on a particular day to gather day-time high temperatures for her
particular area
T1 = 72◦ , T2 = 68◦ , . . . ,
1
ultimately terminating the process with something like T 1826 , one can imagine such data collection
as never-ending and, in this never-ending mindset, the collection of numbers (T n )∞
n=1 is known as a
sequence. Suppose this meterologist has a oceanographer friend in a nearby coastal town. He (the
oceanographer) agrees to begin collecting peak ocean temperatures for each day at a certain point
offshore. If his meterologist friend has been collecting data for a year already, he may wish to label
his measurements starting with the index 366:
O366 = 52◦ , O377 = 51.7◦ , . . . .
Despite the fact that his index does not begin at 1, his process may also be viewed as a sequence,
on-going forever. One might say the domain of her sequence — i.e., the list of indices, or possible
inputs — is the full set of natural numbers N = {1, 2, 3, 4, 5, . . .}, while the domain of his is the
natural numbers from 366 on. One can conceive also of situations where one would wish to start
from a negative index.
Definition 1.1 A sequence is a function whose domain is the collection of integers starting at a
certain integer N0 .
One tends to write things like (an )∞
n=5 , indicating in more compact form that there is a sequence
of values
a5 , a 6 , a 7 , . . .
that is being discussed. Just as permissible, and saying exactly the same thing, would be notations
like
{a(n) | n ≥ 5, an integer}
or
an , n ≥ 5.
The last of these is, of course, not explicit about the fact that n should be an integer. It is a
matter of convention (and, perhaps laziness) that when letters like x, y, z are used as independent
variables, the writer usually wants us to think that they can be numbers from a continuous domain
of inputs, whereas letters like i, j, k, m and n are more often used when the domain is some subset
of the integers. We will follow this convention in what follows.
While the sequences we experience in life may not come from formulas, the ones we study in
mathematics often do. It is in our nature as “scientists” that we study the information that comes
our way, trying to observe patterns and expressing those patterns with succinct, explicit formulas.
The sequence
1
1
a1 = 1, a2 = , a3 = , . . .
2
3
has what is probably an obvious pattern to many, one which may be succinctly summed up in the
explicit formula
1
an = , n ≥ 1.
n
Notice this is really just a function. When we write
f (x) = x2 − 2x, x ∈ [0, 2],
we are expressing a formula and a domain for a function f in such a way that a reader can see at
a glance which inputs are allowable (only those real numbers between 0 and 2 inclusive), and how
to find an output for any particular input of interest. The formula
an =
1
, n ≥ 1,
n
2
achieves precisely the same thing, which should not be surprising since sequences are functions
(look again at the definition above). Because we used n as the independent variable, it is tacitly
understood that the inputs must be integers greater than or equal to 1; placing the input variable
as a subscript an rather than within parentheses a(n) is another subtle indicator that only integer
inputs are expected.
One may ask, “Why do sequences appear in a calculus course?” The more we probe this question,
the more insightful we see it to be. After all, what are the concepts associated with Calculus? The
fundamental calculus concept is that of a limit, upon which we build the concepts of derivative and
integrals (both of which are certain kinds of limits). Now, suppose we have a sequence (a n )∞
n=1 ,
and consider the notion
lim an ,
analog to
lim f (x).
n→3
x→3
You should recall that limits need not exist, but when the limit on the right exists and, say, has
value L, what this means is that values of x arbitarily close to x = 3 are included in the domain
of f (though x = 3 is possibly absent from that domain), and that one can get outputs from f
which are very close to L (within any tolerance level one likes) by restricting ones vision to inputs
x sufficiently close to x = 3. Turning to our sequence a n we see that, if the domain (values of n)
includes only integers, the whole business of a limit as n → 3 is meaningless. One cannot get any
closer to n = 3 than n = 2 or n = 4. While a 3 exists, (just as f (3) exists, if 3 is in the domain
of f ), one simply cannot talk about lim an . A proper understanding of derivatives reveals that, if
n→3
there are no limits at n = 3 (nor at n = any integer), then there can be no derivatives of sequences
either — that is, no such thing as a03 . Given this, the import of the question “why study sequences
in calculus?” should become more clear. Since sequences (sometimes thought of as functions of a
discrete variable n) are more common to our experience than functions of a continuous variable x,
perhaps the question should be turned around to ask, “Now that we have learned what the concepts
of calculus can tell us about functions of continuous variables, how can we use this knowledge on
the kinds of functions (sequences) we experience in everyday life?”
In fact, no direct attempt will be made to answer the previous question, though it is indicative
of the mindset behind the study of sequences. We hope that, as they are studied, you will look for
ways to see the subject of sequences along with all the material we have studied prior to it in a
coherent fashion. One way to apply prior concepts is in a question like this one: “So, I have this
sequence of numbers. Where is it going? Is it settling into a limit?” Indeed, we do not contradict
what was said before about taking limits with sequences if we consider the notion
lim an
n→∞
analog to that of
lim f (x),
x→∞
for one can get “as close to infinity” considering only integer values n as one can get considering
real values x. Indeed, it would seem that the sequence
1 1 1
1
which consists of the numbers
, , , ...
an = , n ≥ 2
n
2 3 4
is tending to zero (or has limit zero) as n → ∞ much in the same way that
1
f (x) = → 0 as x → ∞.
x
This is perhaps illustrated more clearly by looking at the graphs, both of the sequence a n = 1/n
and of the function f (x) = 1/x.
3
an
fHxL
1
1
2
4
6
n
8
2
4
6
8
x
We may superimpose the two graphs, which looks like this:
fHxL
1
2
4
6
8
x
So, we make the following definitions, based on the idea that the only limit one can discuss concerning a sequence is its limit as n → ∞. The first is precisely the definition of a “limit of a function
at infinity” applied to sequences.
Definition 1.2 We write lim an = L if, for each number > 0 there exists a corresponding
n→∞
integer N such that
|an − L| < whenever
n ≥ N.
Definition 1.3 The limit of a sequence a n is the value lim an , if this limit exists. If the limit
n→∞
exists the sequence is said to be convergent. Otherwise, it is said to be divergent.
Of course, if there is even one tolerance level > 0 for which, no matter how far out in the
sequence you go (i.e., no matter how large of an N you choose), there are still values a n for n ≥ N
which are farther than from L (a particular number), then L is not the limit of the sequence.
And, if no L exists which is the limit of a n , then we say lim an does not exist.
n→∞
How does one find the limit of a sequence? The figure above in which we superimposed the
graphs of an = 1/n and f (x) = 1/x suggests one possible method — to find a function of a
continuous variable which “connects the dots” and assess its limit at infinity. Often it will be very
easy to think of a function f (x) which connects the dots. It was most natural function to think of
f (x) = 1/x when thinking of connecting the dots in the graph of a n = 1/n. Having thought of it,
we can say
1
1
lim
= lim
= 0.
n→∞ n
x→∞ x
Similarly, for the sequence
2n2 − 7
an =
3n − n2
4
we think of f (x) = (2x2 − 7)/(3x − x2 ) and assess the limit of the sequence in the same fashion
2n2 − 7
n→∞ 3n − n2
lim
2x2 − 7
x→∞ 3x − x2
4x
= lim
(by L’Hôpital’s Rule)
x→∞ −2x
= lim (−2)
(cancelling the common factor 2x)
=
lim
x→∞
= −2.
Keep in mind that, in order to use L’Hôpital’s Rule, we had to make the switch over to a function
of a continuous variable x, for L’Hôpital’s Rule requires a numerator and denominator which are
differentiable. The numerator and denominator of a n = (2n2 − 7)/(3n − n2 ) may both be thought
of as sequences, each of which has no derivative. (Recall our remarks above about differentiating
sequences.) This is a subtle but important point. Though it may seem just as good (and more
efficient) to write
4n
2n2 − 7
=
lim
(by L’Hôpital’s Rule)
lim
n→∞ −2n
n→∞ 3n − n2
= lim (−2)
(cancelling the common factor 2n)
n→∞
= −2,
you should only do so with the deeper understanding described above — i.e., that we are passing
from a sequence over to a function of a continuous variable in the process (as, in this case, was
required in order to use L’Hôpital’s Rule).
In the previous two examples, we took the most obvious (and thereby the most convenient)
functions f (x) that did the job of “connecting the dots” on the graph of the sequence — f (x) = 1/x
for the sequence 1/n, and f (x) = (2x2 −7)/(3x−x2 ) for the sequence (2n2 −7)/(3n−n2 ). For some
sequences, however, the most obvious candidate doesn’t actually make sense. One such sequence is
an = (−1)n , n ≥ 0.
If one does not think carefully about it, one may think
f (x) = (−1)x , x ≥ 0
an appropriate function of a continuous variable x to connect the dots on the graph of a n . But, keep
in mind what this must mean: that we have extended the original function (sequence) a n , defined
for all integer values n = 0, 1, 2, 3, . . ., to a function f (x) defined for all real values of x ≥ 0. Now
look closely at the formula we wrote down for f . It is not defined at x = 1/2, at least, not unless
we are willing to accept non-real outputs (in this case, the imaginary number i). We have been
judiciously avoiding complex numbers in calculus and, unless we wish to start the whole calculus
sequence over with functions f (x) which can have non-real outputs (something that can be and is
done in Math 365, but only with students who have more mathematical experience), we had better
not use this formula for f . One might quite validly observe that it is not really necessary that our
f connect up all of the dots, so long as it connects all of the from a certain point onward. But
proposing
f (x) = (−1)x , x ≥ 100
5
is no improvement for, once again, f (100.5) is not defined.
Granted, there are other formulas, other functions which connect up the dots and do not slip
over to non-real outputs between integer inputs, but finding such a formula is more work than we
really need to do if we realize that there are other valid ways to find the limit of a sequence besides
the method of extending to a function of a continuous variable. Perhaps the most important other
method, and one that works with our sequence a n = (−1)n , is to consider subsequences. Two
convenient subsequences of
an = (−1)n , n ≥ 0 :
1, −1, 1, −1, . . .
are the subsequence
1, 1, 1, 1, . . . ,
which are the terms a0 , a2 , a4 , . . . of the sequence, and the subsequence
−1, −1, −1, −1, . . . ,
which are the terms a1 , a3 , a5 , . . .. Both of these subsequences are sequences in their own right.
Looking at the graph of the full sequence a n = (−1)n , we see that the first subsequence is represented by the black dots and the second by the gray ones.
1
0.5
2
4
6
8
10
-0.5
-1
If the full sequence an = (−1)n had a limit, it seems reasonable to expect that each subsequence of
an would have a limit as well — in fact, have the same limit as the full sequence. This expectation
is, indeed, correct, and can be stated more generally as the theorem:
Theorem 1.1 For any convergent sequence with limit L, it is the case that every subsequence also
has limit L.
It is clear that the subsequence a0 , a2 , a4 , . . . has limit 1, while the subsequence a 1 , a3 , a5 , . . .
has limit (−1). This is incompatible with the notion that a n = (−1)n has a limit, for if it did, all
of its subsequences would have the same limit. Thus, we conclude that (−1) n is divergent.
2
Geometric Series
The title suggests the main topic of this section is geometric series but, in fact, this is the second of
two main topics. The first is the concept of series, or infinite series (which is what mathematicians
really mean whenever they talk about series). Each infinite series has an underlying sequence with
6
which it is associated, and that is why this topic must come after the previous section. Give the
sequence
an = (−1)n ,
the associated series is
∞
X
n=0
(−1)n = 1 − 1 + 1 − 1 + 1 − 1 + . . . .
The idea of an infinite sum is, perhaps, a strange concept, though if you consider a bit what we
were doing when taking integrals, were they not infinite sums? For any sequence a n , n ≥ N0 , there
is an associated infinite series
∞
X
an = aN0 + a1+N0 + a2+N0 + . . . .
n=N0
One rightly may be skeptical about whether such an infinite series (sum) converges (i.e., has a
numeric answer in the limit). Indeed, the sequence a n = 1, n > 0 has the associated series
∞
X
1 = 1 + 1 + 1 + 1 + ...
n=1
which (as any reasonable person would correctly guess) equals +∞, not a finite number. One
might, however, be less certain about the infinite series
∞
X
n=0
(−1)n = 1 − 1 + 1 − 1 + 1 − 1 + . . . .
Viewed one way, it would appear the sum might be zero:
(1 − 1) + (1 − 1) + (1 − 1) + . . . = 0 + 0 + 0 + . . . = 0.
But viewed with the terms grouped differently, it would appear the sum might be one:
1 + (−1 + 1) + (−1 + 1) + (−1 + 1) + . . . = 1 + 0 + 0 + 0 + . . . = 1.
It is not possible for both answers to be correct, and we need a better understanding of series if we
are to assess which (if, in fact, either one) is correct.
The correct way to understand a series is to see it as the limit of a sequence, a secondary
sequence built from the original sequence with which it is associated. If we start with a sequence
an , n > 0, we may build a new sequence sn , n > 0 from it in the following way: we set
s1 = a 1 ,
s2 = a 1 + a 2 ,
s3 = a 1 + a 2 + a 3 ,
..
.
sn = a 1 + a 2 + · · · + a n =
..
.
7
n
X
j=1
aj ,
∞
The new sequence (sn )∞
n=1 is known as the sequence of partial sums of the sequence (a n )n=1 . Of
course, armed with this new sequence (s n )∞
n=1 , we may repeat the process to build a sequence of
partial sums of the sequence of partial sums, but that is a side issue we will not explore. The main
issue is this: that the infinite series
∞
X
j=1
aj = a 1 + a 2 + a 3 + · · ·
is precisely what one investigates when one considers the limit of the sequence of partial sums:
lim sn .
n→∞
Let us be more concrete with an example. Consider once again the series associated with the
sequence an = 1, n > 0. Its sequence of partial sums is
s1 = 1,
s2 = 1 + 1 = 2,
s3 = 1 + 1 + 1 = (3)(1) = 3,
..
.
sn = (n)(1) = n,
..
.
The explicit formula for the nth term in the sequence of partial sums is s n = n, which clearly has
limit +∞ as n → ∞. This is just what (intuitively) we felt this infinite sum would be. Let’s apply
the same kind of analysis to the series associated with a n = (−1)n , n ≥ 0. Its sequence of partial
sums is
s0 = 1,
s1 = 1 − 1 = 0,
s2 = 1 − 1 + 1 = 1
s3 = 1 − 1 + 1 − 1 = 0
..
. (
1, if n is even
sn =
0, if n is odd,
..
.
The graph of (sn )∞
n=0 looks like this
1
0.8
0.6
0.4
0.2
2
4
6
8
10
and should remind you of the sequence we analyzed at the end of the last section. Just as with
that sequence, there are two natural subsequences of (s n )∞
n=0 which have different limits, and this
8
leads us to the conclusion that the sequence of partial sums is divergent (and hence so is the infinite
P
n
series ∞
n=0 (−1) ). So, neither of the answers (0 and 1) we proposed originally for this series is
correct.
Now let us turn our attention to sequences of a particular form a n = r n , n ≥ 0, where r is
P
n
any real number, and the infinite series ∞
n=0 r associated with such sequences. These series are
collectively classified as being geometric. Analyzing this series as before, we look at the sequence
of partial sums:
s0 = 1,
s1 = 1 + r
s2 = 1 + r + r 2
..
.
sn = 1 + r + r 2 + · · · + r n
..
.
In the previous two instances of series we analyzed, we obtained nice closed-form expressions for
the nth term in the sequence of partial sums: s n = n in the first case, and sn = 1 or 0 in the
second, depending upon whether n was even or odd. If our analysis here is to be fruitful, we should
hope for the same thing — a nicer formula for s n than what we currently have. There is a clever
method for getting it. If we write down what we currently have for s n and, just below that, the
same equation multiplied through by r, we get
sn = 1 + r + r 2 + · · · + r n
rsn =
r + r 2 + · · · + r n + r n+1
−−
−−−−−−−−−−−−−
(1 − r)sn = 1 − r n+1 ,
where this last equation arises by subtracting the two equations above it. Solving the last equation
for sn gives us
sn =
1 − r n+1
.
1−r
(1)
P
n
As before, we assess whether or not the series ∞
n=0 r converges by determining whether lim n sn
exists. If r = 1 this expression for s n is meaningless (0/0), but a look back to the original series
P∞ n
n=0 r reveals that, if we take r = 1, then we are dealing with one of our previous examples
which diverged. So, let us turn to the cases in which r 6= 1. Since there is only one place in this
formula for sn in which n appears, we may write
1 lim sn =
1 − lim r n+1 ,
n
n
1−r
with the understanding that the limit on the left exists if and only if the limit on the right (lim n r n+1 )
exists. For r > 1, the limit on the right does not exist, and the series diverges. For 0 ≤ r < 1, the
limit on the right is zero, in which case we get
∞
X
n=0
r n = lim sn =
n
1
.
1−r
(2)
It is not quite so obvious what happens for r < 0, although the other one of the previous examples
we handled (also divergent) corresponds to the case r = −1. In fact, when −1 < r < 1 the graph
9
of the sequence an = r n looks like this
2
4
6
8
10
giving the appearance (correct, and proved later) that r n (and hence r n+1 ) goes to zero as n → ∞.
Thus, formula (2) holds for −1 < r < 0 as well. When r < −1, the graph of a n = r n looks like this
1
2
4
6
8
10
Again, we see two subsequences, one going to +∞ and the other to −∞. Thus, lim n r n+1 does not
exist for r < −1. We summarize these results in this statement: The geometric series
∞
X
n=0
rn =
1
,
1−r
if |r| < 1
(3)
(that is, it converges to a known value), and diverges if |r| ≥ 1.
One thing to note is that the geometric series whose sum we have found always starts with the
term r 0 = 1. If we wish to find sums of other geometric series, we may have to do a little work.
For instance,
#
" 6 7
n
∞
5
X
1
1
1
3
1
3
· (−1)n
=
+
−
+ ···
−
4
7
4
7
7
7
n=5
#
5 "
2 3
3
1
1
1
1
= −
1− +
−
+ ···
4
7
7
7
7
5
3
1
1
= −
·
(since r = −1/7 is between -1 and 1)
4
7
1 + 1/7
5 1
7
3
= −
4
7
8
3
= −
.
76832
Incidentally, formula (1) for the partial sums of a geometric series, which we used to take the
limit of such partial sums, is valid regardless of whether |r| < 1, and is immensely useful in its own
right. For instance, suppose you are considering a purchase (a house, a car, a college education) for
which you will have to borrow P0 dollars at annual interest rate I. Your monthly payments will be
10
m dollars, and you will have to make N such payments over the life of the loan. Let us denote by
Pn the balance of the loan after you’ve made your nth payment. As your first payment is usually
due one month after taking out the loan, we can write down a formula for the amount you owe
after making that payment as:
I
− m.
P1 = P 0 1 +
12
Similarly,
P2 =
=
=
P2 =
=
..
.
I
P1 1 +
−m
12
I
I
P0 1 +
−m 1+
−m
12
12
I
I 2
P0 1 +
−m 1+ 1+
12
12
I
P2 1 +
− m = · · · (algebra steps omitted)
12
"
#
I 2
I
I 3
P0 1 +
+ 1+
−m 1+ 1+
12
12
12
Pn = P 0
I
1+
12
n
"
I
−m 1+ 1+
12
#
I 2
I n−1
+ 1+
+ ··· + 1 +
.
12
12
The quantity inside the square brackets in our expression for P n is a partial sum of a geometric
series. Employing formula (1), we have
1 − (1 + I/12)n
I n
−m
,
Pn = P 0 1 +
12
1 − (1 + I/12)
or
Pn = P 0
I
1+
12
n
+
12m
[1 − (1 + I/12)n ].
I
(4)
This formula may be used in several ways. The way it appears here, you could use it directly to
determine how much you would still owe after 12 months (i.e., to compute P 12 ). Or, if you wanted
to make a large enough payment m so that P 24 was some fixed amount, you could stick in that
amount (also put in n = 24) and solve for m. The author has surprised several mortgage agents by
being able to calculate numbers such as these.
As we close this section, we bring up one issue that the insightful reader may have already
suspected. In the section we dealt with three separate series (the first two important special cases
of the third):
(1) 1 + 1 + 1 + 1 + · · · ,
(2) 1 − 1 + 1 − 1 + · · · ,
(3) 1 + r + r 2 + r 3 + · · · ,
11
each of which we analyzed by finding a closed-form expression for the nth partial sum:
(1) sn = n,
(2) sn =
(
(3) sn =
1 − r n+1
.
1−r
1, if n is even,
0, if n is odd,
Happening upon such nice formulas for the nth partial sum is the exception, not the rule. For most
series, it is not possible to get a closed-form expression for s n , in which case we cannot easily assess
whether they converge or diverge through considering lim n sn .
3
The Divergence Test and p-series
As the title suggests, there are two separate topics to which this section is devoted.
3.1
The Divergence Test
One thing we may have noted from our study of geometric series
∞
X
n=N0
r n = r N0 + r 1+N0 + r 2+N0 + · · ·
is that the series converges precisely in those cases when the nth term r n−1 goes to zero as n → ∞,
and diverges when this same nth term does not go to zero as n → ∞. One might ask whether this
is a general rule that all convergent and divergent series follow. Let us first suppose that we have
P
a convergent series ∞
n=1 an . (We take the starting index to be 1 for convenience, not because it
has to be so for the analysis that follows.) Thus lim n sn exists, where (sn )∞
n=1 is the sequence of
partial sums give by
s1 = a 1
s2 = a 1 + a 2
..
.
sn = a 1 + a 2 + · · · + a n
..
.
Since limn sn exists, so does limn sn−1 ; in fact, both of these limits have the same value, which
means that
lim an = lim (sn − sn−1 )
n
n
= lim sn − lim sn−1
n
n
= 0.
We have proved the following theorem.
Theorem 3.1 If the series
P∞
n=N0
an converges, then limn an = 0.
12
Many mathematical theorems are stated in the logical form
If p then q,
P
where p and q are assertions. In the above theorem, p is the assertion that “ ∞
n=N0 an converges”,
whereas q is the assertion that “lim n an = 0”. Any two assertions p and q can be arranged in an
“if p then q” sentence. In fact, there are four such arrangements which logicians usually consider:
If
If
If
If
p then q. (Let’s call this the “base” statement.)
q then p (called the converse).
not q then not p (called the contrapositive).
not p then not q (the contrapositive of the converse).
Of course, not all “If p then q” sentences are to be summarily believed. Just think of a nonsense
sentence like, “If he has red hair then fish can fly”. Even when such a sentence is unilaterally true,
not all of its variants are likewise true. Consider the statement
If it is a star, then it is hot.
Assuming that this statement is to be believed, only one of its three variants is also to be believed:
If it is hot, then it is a star. (This is the converse, and it is not true.)
If it is not hot, then it is not a star. (The contrapositive is true.)
If it is not a star, then it is not hot. (The contrapositive of the converse is
false.)
Just because a certain “If p then q” statement is true does not mean that the converse is true as
well. (In the example above, the converse is not true. However, taking the different statement, “If
he is the 16th president of the U. S., then he is Abraham Lincoln”, one gets a case where both
the base statement and its converse are true.) However, given any true “If p then q” statement,
its contrapositive is true as well. Applying this to our last theorem, the truth of which has been
proved, we get another theorem which is the contrapositive.
Theorem 3.2 (Divergence Test) If lim n an 6= 0 (that is, if this limit is some other value than
P
zero or if it does not exist), then the series ∞
n=N0 an diverges.
Theorem 3.2 says that a necessary condition for convergence of a series is that its nth term go
to zero as n → ∞, a statement that often proves more useful than the original statement (Theorem
3.1). For instance, it tells us at a glance that the series
∞
X
2n2 − 7
3n − n2
n=4
diverges. (Recall that in Section 1 we showed that lim n (2n2 − 7)/(3n − n2 ) = −2, not 0.) What
we do not yet know is whether or not the converse statement of Theorem 3.1 is true; that is, we
do not know if limn an = 0 is a sufficient condition for convergence. We will get the answer to this
question in our discussion of p-series.
13
3.2
The p-series
In Section 2 on geometric series we analyzed series of the form
∞
X
n=N0
r n = r N0 + r 1+N0 + r 2+N0 + · · · ;
that is, series whose terms are all the same constant r raised to powers which grow with n. In this
section we analyze the so-called p-series, which are series of the form
∞
X
n=N0
np = N0p + (1 + N0 )p + (2 + N0 )p + · · · ;
that is, ones with a fixed exponent and a base that grows with n. Conclusions we drew for geometric
series do not help us in the analysis of p-series, and it is important to train yourself to tell the one
type of series from the other (much as, when dealing with exponential functions f (x) = a x in Math
161, you learned a different rule for finding f 0 (x) than for the power functions f (x) = x n ).
For convenience sake, we will once again take the starting index N 0 to be 1, giving us p-series
of the form
∞
X
n=1
np = 1 + 2 p + 3 p + 4 p + · · · ,
(5)
though none of the conclusions we draw concerning such series relies on their starting at n = 1 (and
after reading through the section, you should go back and think about how the arguments change
if we start, say, at n = 5 or any other positive integer). Unlike geometric series, there is no nice
closed-form expression for the nth partial sum of a p-series, so we take a different approach than
that taken in Section 2. We will first use the Divergence Test to rule out convergence whenever
the nth term of the series does not go to zero. The nth term of the series (different from the nth
partial sum) is an = np , a power whose base grows from one term to the next. Check that the
following three statements are true:


 0, if p < 0,
p
lim n =
1, if p = 0,
n


+∞, if p > 0.
By the Divergence Test, we have determined that the p-series (5) diverge when p ≥ 0.
We still do not know about the case p < 0. (Recall that we have no theorem stating that
P∞
limn an = 0 is a sufficient condition for the convergence of the series
n=0 an .) Let us take a
graphical approach to the question. We may represent each term in the sum (5) as the area of a
rectangle — that is, we will draw a rectangle R 1 whose base is 1 and whose height is a1 = 1 (so its
area is a1 ), a second rectangle R2 whose base and height are 1 and a2 = 2p respectively, and so on.
Arranged side-by-side, these rectangles might look like this:
14
1
R1
R2
R3
R4
Viewed in this way, the sum of the series (5) is the sum of the areas of these infinitely-many rectangles. We cannot easily compute this sum (if, indeed, it converges), but by comparing this sum of
areas of rectangles to other areas, we can find out whether the sum converges or not. To do this,
we note that if the rectangle R1 is placed so that its bottom left corner is positioned at x = 0, then
the graph of f (x) = xp conveniently remains above every rectangle, touching each one at its top
right corner
1
2
4
6
8
10
whereas if we slide these rectangles one unit to the right, then this same curve intersects each at
its top left corner
1
2
4
From these observations we conclude that
Z
p
p
p
2 + 3 + 4 + ··· ≤
or
∞
X
n=2
np ≤
∞
1
Z
R∞
6
10
xp dx ≤ 1 + 2p + 3p + 4p + · · · ,
∞
1
8
xp dx ≤
xp dx
∞
X
np .
n=1
But we have studied integrals of the type 1
and have found that they diverge when p ≥ −1
and converge when p < −1 (i.e., there is infinite area under the curve x p when p ≥ −1 and finite
15
area when p < −1). In the case −1 ≤ p < 0, we use the right inequality to conclude that
∞
X
np
n=1
diverges. In the case p < −1, the left inequality tells us that the sum (as well as its partial sums)
∞
X
np
n=2
has an upper bound (i.e., it cannot be infinite). This, by itself, would not be enough to know that
this sum converges. We saw an illustration of this statement when working with the sum
1 + (−1) + 1 + (−1) + 1 + (−1) + · · · ,
(6)
whose partial sums swing back-and-forth between 0 and 1; 1 is an upper bound on the sequence of
partial sums, but the series (6) diverges. What is different about the sum
2p + 3 p + 4 p + 5 p + · · ·
is that each of its terms np are positive. The partial sum sn+1 is always bigger than its predecessor
sn . There is a theorem (beyond our level to prove) which says
Theorem 3.3 All bounded, non-decreasing sequences of (real) numbers converge.
This theorem applies directly to the sequence of partial sums
s2 = 2 p ,
s3 = 2p + 3p ,
s4 = 2p + 3p + 4p ,
...,
each term of which is larger than its predecessor and is bounded above by the convergent integral
R∞ p
1 x dx (convergent because p < −1). Hence, the sequence
∞
X
np
n=2
converges for p < −1, which means
converges as well (since its value is only one larger than
the following answers to questions posed previously:
1. The p-series
when p ≥ 1.
P∞
n=1
np (or, more generally,
P∞
P∞
n=N0
n=2
∞
X
np
n=1
np ). Therefore, we have determined
np ) converge when p < −1 and diverge
2. The condition limn an = 0 is not sufficient for convergence of the series
P∞
n=N0
an .
The borderline case between convergence and divergence of the p-series is the case p = −1
∞
X
n=1
n−1 = 1 +
1 1 1
+ + +···,
2 3 4
which we know to be divergent. This is a series studied since the time of the ancient Greeks, and
is known as the harmonic series.
16
4
Absolute and conditional convergence; alternating series
In Section 3.2, the terms an of the series we discussed were all positive. In fact, throughout
our study of series thus far, we have only encountered series containing negative terms in certain
geometric series cases, namely those in which r < 0, and in these cases the negative terms appear
in a quite predictable way (every other term). The behavior of series containing negative terms —
even those in which the negative terms appear in a predictable fashion — can, in general, be quite
strange. This is in contrast to series whose terms are all nonnegative for, as we saw at the end
of the previous section, a consequence of Theorem 3.3 is that such series can do only one of two
things: diverge to infinity, or converge. Of course, if all of the terms of a series are negative, then
the negative may be factored out, and we are back to a series with positive terms. So, the kinds
of situations which may pose new difficulties for us are those in which the series has both positive
and negative terms — in fact, infinitely many of each.
One tool for investigating the convergence of a series which has negative and positive terms —
a series that is not geometric — comes from the following theorem, which will not be proved here.
Theorem 4.1 If
P∞
n=N0
|an | is convergent, then so is
P∞
n=N0
an .
Theorem 4.1 provides a sufficient but not necessary condition for convergence of a series. Given
P∞
a series
n=N0 an with both positive and negative terms, one thing you might try in order to
P
determine if the series converges is to see if the series ∞
n=N0 |an | — i.e., the same series with all
of the minus signs dropped — converges. If it does, this is sufficient (by the theorem) to conclude
P
that ∞
n=N0 an converges, too. If it does not, no conclusion can be drawn — that is, convergence
of the latter is not necessary for convergence of the former. We already know that a series can
either converge or diverge. The above theorem suggests we may make a finer distinction within the
category of convergent sequences.
P∞
P∞
Definition 4.1 A series
n=N0 |an | converges.
n=N0 an is said to be absolutely convergent if
P∞
P∞
(Note: by Theorem 4.1 we know in this case that
n=N0 an converges as well.) If
n=N0 |an |
P∞
P∞
diverges but n=N0 an converges, then the series n=N0 an is said to be conditionally convergent.
Here are some facts to go along with these definitions:
P
• A divergent series ∞
n=N0 an cannot be convergent in any of the senses mentioned.
P
P∞
• A convergent series ∞
n=N0 an may have an associated series
n=N0 |an | which is divergent.
In that case, the former is conditionally convergent.
P∞
P
• A convergent series ∞
n=N0 |an | which is convergent.
n=N0 an may have an associated series
In that case, the former is absolutely convergent.
As an example, suppose we are interested in knowing whether the series
1−
1 1
1
1
1
1
1
1
+ −
−
+
−
−
−
+ ···
4 9 16 25 36 49 64 81
converges. Theorem 4.1 tells us we can make the following attempt: drop all of the minus signs
and see if
1
1
1
1
1
1
1 1
+
+
+
+
+
+ ···
1+ + +
4 9 16 25 36 49 64 81
17
P∞ −2
converges. In fact, this is the series
n=1 n , which is a convergent p-series, so the original
converges (absolutely).
At this point you may rightly ask, “what if I am given a series which I find does not converge
absolutely? How do I determine if it converges conditionally?” This question is easily answered if
the terms of the series are all positive or all negative (Do you see why?), but can be quite difficult
in the general case. We will not attempt to give a complete answer — indeed, I doubt a complete
answer is available. But, if the signs of the terms alternate back and forth term-by-term (called an
alternating series) as in
∞
X
(−1)n−1
n=1
n
= 1−
1 1 1 1
+ − + − ···,
2 3 4 5
there is a test (theorem) which is sometimes helpful.
Theorem 4.2 (Alternating Series Test) Suppose that a n > 0 for each n = 1, 2, 3, . . ., and that
these terms additionally satisfy the conditions:
(i) an+1 ≤ an for all n ≥ N (i.e., at some point along the way the terms never again increase),
and
(ii) limn an = 0.
Then the alternating series
∞
X
n=1
(−1)n−1 an = a1 − a2 + a3 − a4 + a5 − a6 + · · ·
converges.
Proof: As with any series, we may try to show that the sequence of partial sums
s1 = a 1 ,
s2 = a1 − a2 ,
s3 = a1 − a2 + a3 ,
...
converges. Consider first the even-numbered partial sums, those of the form
s2n = a1 − a2 + a3 − a4 + · · · − a2n .
Notice that, for two consecutive even-numbered partial sums, their difference
s2n+2 − s2n = a2n+1 − a2n+2 ≥ 0,
by condition (i). So, within the sequence (s n )∞
n=1 of partial sums there is a subsequence
(consisting of the even-numbered ones)
s2 , s 4 , s 6 , s 8 , , . . .
18
which is non-decreasing. As such, this subsequence either diverges to infinity or converges. To see that it converges, we need only show that there is some upper bound to
each of the s2n ’s (by Theorem 3.3). But, an upper bound is provided by
s2n = a1 − a2 + a3 − a4 + · · · − a2n
= a1 − (a2 − a3 ) − (a4 − a5 ) + · · · − (a2n−2 − a2n−1 ) − a2n
≤ a1 − a2n
≤ a1
(by condition (i))
(since each term is nonnegative),
since each of the quantities in parentheses is nonnegative (again by (i)). Since bounded
(by a1 ) and non-decreasing, the sequence s 2 , s4 , s6 , . . . converges to a number s. Considering a typical odd-numbered sum s 2n+1 , we have
lim s2n+1 = lim(s2n + a2n+1 ) = lim s2n + lim a2n+1 = s (by condition (ii)).
n
n
n
n
That is, the odd-numbered sums go to the same limit as the even-numbered ones, and
hence the full sequence (sn )∞
n=1 of partial sums has a limit, which is what we set out to
prove.
Notice that
• The Alternating Series test applies directly to our alternating series
1−
1 1 1 1 1
+ − + − + ···,
2 3 4 5 6
telling us that it converges. (Verify for yourself that all of the hypotheses of the test are in
place.) It applies more generally to alternating geometric series with −1 < r < 0:
∞
X
n=0
rn =
∞
X
n=0
(−1)n |r|n = 1 − |r| + |r|2 − |r|3 + · · · .
You may recall that when we handled this interval of r values in Section 2, our proof that
the series converged relied on our trusting the appearance of a graph.
• The test applies to non-alternating series as well, so long as the terms eventually satisfy the
hypotheses. That is, it applies if, after having thrown away the first 100 or 1000 (or however
many necessary) terms, the ones that remain satisfy the hypotheses.
5
5.1
Power series
Power series; the Root and Ratio Tests
Up until now the series we have viewed have been series of numbers. Consider now the following
definition.
Definition 5.1 A series of the form
∞
X
n=0
is called a power series centered at a.
cn (x − a)n
19
Notice that a series such as this is really a function. The domain of this function consists of those
x-values for which the series converges, and is called the interval of convergence. The reason for
this name becomes clear by the next theorem
Theorem 5.1 Let (cn )∞
n=0 be a sequence of numbers, giving rise to a power series
∞
X
n=0
centered at a. One of the following holds:
cn (x − a)n
(i) The power series converges at x = a, but diverges at every other value of x.
(ii) The power series converges for all real numbers x.
(iii) There is a positive number R such that the series converges whenever |x − a| < R (that is,
for all numbers less than R units from a) but diverges whenever |x − a| > R.
Notice that, in each instance listed above, x = a is in the middle of the domain of the power series.
The number R in (iii) is called the radius of convergence, as it indicates just how far one may stray
from the center at a (in either direction) before one encounters divergence of the series. Convention
is that, in case (i), one writes R = 0 and in case (ii) R = +∞.
Examples do not prove the truth of general statements such as the one made in the above
theorem. Still, since we will not provide a proof of this theorem, it is nice to see that such general
statements apply to examples we already have seen. Notice that the geometric series (writing x
instead of r)
∞
X
xn
n=0
converges for |x| < 1 — that is, it is a power series (each c n = 1, for n = 0, 1, 2, . . .) centered at 0
with radius of convergence R = 1. Its interval of convergence is −1 < x < 1, or (−1, 1).
It should be relatively easy, given a particular power series, to determine where it is centered.
To find the radius of convergence, one of the following tests is usually employed.
∞
X
an+1
= ρ. The series
an
Theorem 5.2 (The Ratio Test) Suppose a n > 0 for each n with lim
n
an
n=0
(i) converges, if ρ < 1, and
(ii) diverges, if ρ > 1.
If ρ = 1, the test is inconclusive.
Theorem 5.3 (The Root Test) Suppose a n ≥ 0 for each n with lim
n
(i) converges, if ρ < 1, and
(ii) diverges, if ρ > 1.
If ρ = 1, the test is inconclusive.
20
√
n
an = ρ. The series
∞
X
n=0
an
Here are some examples of the use of these two theorems in the determination of radii of
convergence. Consider the power series
∞
X
1 · 3 · 5 · · · (2n + 1)
n!
n=0
(x − 2)n .
This is a power series centered at x = 2 whose nth term is a n = 2n (x − 2)n /n!. Notice two things:
• We have the freedom to try either/both of the previous tests. Nevertheless, both tests call for
series whose terms are nonnegative, and because of the presence of (x − 2) in the expression
for an , we know that some terms will be negative if x < 2 (i.e., a value to the left of the center
of the series). Because of Theorem 5.1, however, this is a non-issue. For any distance from
center up to (but not including) the radius of convergence, convergence will occur whether x
is to the left or to the right of center. Thus, we may consider |a n | which is positive instead of
an .
• In deciding which test to try first, we may consider which is easier between taking the limit
of the quantities
|an+1 |
|an |
and
=
p
n
|an | =
1 · 3 · · · [2(n + 1) + 1]|x − 2|n+1 /(n + 1)!
1 · 3 · · · (2n + 1)|x − 2|n /n!
p
n
1 · 3 · · · (2n + 1)|x − 2|n /n!.
The first (the ratio test) option appears to lead to a lot of cancellation, so going with this we get
lim
n
|an+1 |
|an |
1 · 3 · · · [2(n + 1) + 1]|x − 2|n+1 /(n + 1)!
n
1 · 3 · · · (2n + 1)|x − 2|n /n!
(2n + 3)|x − 2|n!
= lim
n
(n + 1)n!
2n + 3
= (|x − 2|) lim
n n+1
= 2|x − 2|.
= lim
For this limit ρ = 2|x − 2| to be less than 1, we need |x − 2| < 1/2 — that is, the radius of
convergence R = 1/2.
Consider now the geometric series centered at x = −3
∞ X
x+3 n
n=0
5
.
Although we have other means for determining the radius of convergence of this series, let us look
at what the root test tells us. Here a n = [(x + 3)/5]n which, once again, need not be positive.
Nevertheless, if we look at the limit
r
n
|x + 3|
|x + 3|
n |x + 3|
= lim
=
.
lim
n
n
n
5
5
5
For this limit to be less than 1 we requre |x + 3| < 1/5, so the radius of convergence R = 5.
21
One can determine most of the details about the interval of convergence by knowing both the
center and the radius. The remaining question is whether there is convergence at the endpoints of
the interval. Consider the question of the interval of convergence for the geometric series
∞
X
(2x)n .
n=0
The series is centered at 0, and an application (similar to the example above) of the root test
reveals a radius of convergence R = 1/2. (You should verify these statements.) This means that
the interval of convergence is one of the following four possiblities:
(−1/2, 1/2),
[−1/2, 1/2),
(−1/2, 1/2],
or
[−1/2, 1/2].
To decide which is correct requires our determining which, if any, of the endpoints x = ±1/2 yields
series convergence. At x = −1/2 we have the series
∞
X
n=0
[2 · (−1/2)]
n
=
∞
X
n=0
(−1)n = 1 − 1 + 1 − 1 + 1 − 1 + · · · ,
which we analyzed in Section 2 and found to be divergent. Similarly, at x = 1/2 we have the series
∞
X
n=0
[2 · (1/2)] n =
∞
X
n=0
1 = 1 + 1 + 1 + 1 + 1 + 1 + ···,
also divergent. Therefore, the interval of convergence for the original power series is (−1/2, 1/2)
(i.e., neither endpoint is included). A general comment here is that neither the root nor ratio
tests are helpful in determining whether a specific endpoint should be included, since it is at the
endpoints where one gets ρ = 1 (the indeterminate case); analysis of endpoints requires the use of
one of our earlier methods
5.2
Power series from other power series
The real value of power series lies in their provision of alternative expressions for functions. In
Section 2 we worked out that
∞
X
1
xn ,
=
1−x
n=0
−1 < x < 1.
(7)
Though the function (1 − x)−1 may seem the less complicated of these two formulas for the same
function (they are the same, at least, in −1 < x < 1), this is the first of potentially many other
functions for which we have alternative expressions as “infinite polynomials”. For certain situations,
we may be satisfied with the approximation to (1 − x) −1 provided by the “truncated series”
N
X
1
≈
xn = 1 + x + x 2 + x 3 + · · · + x N ;
1−x
n=1
that is, the partial sum of the series that stops at the N th power of x. While (1 − x) −1 may
be simpler than its full power series, it is harder to argue that it is simpler than an N th-degree
polynomial.
22
At any rate, whether or not you buy the argument of the previous paragraph for the study
of power series as means to approximating functions, what we wish to do now is to use power
series such as equation (7) to get power series expressions for other functions. One way is through
algebraic manipulation, which may be done if the function with which we are working is similar
enough to one for which we already know a power series expansion. Take, for instance, the function
f (x) = (1 + x)−1 . This function looks somewhat similar to (1 − x) −1 , and by making this similarity
clear, we can get a power series expansion for f :
1
1+x
1
1 − (−x)
∞
X
(−x)n
=
=
=
n=0
∞
X
(−1)n xn ,
n=0
−1 < x < 1.
The interval of convergence is derived from that for the series (7) by noting that −1 < −x < 1
implies −1 < x < 1.
Similarly, for (1 + x2 )−1 ,
1
1 + x2
1
1 − (−x2 )
∞
X
(−x2 )n
=
=
=
n=0
∞
X
(−1)n x2n ,
n=0
−1 < x < 1.
The interval of convergence is derived from that for the series (7) by noting that −1 < x 2 < 1
implies −1 < x < 1.
A more complicated, yet similar example is the power series for 3x/(2 − x):
3x
2−x
3x
1
·
2
1 − x/2
∞
X x n
= 3x
2
=
= 3x
n=0
∞
X
2−n xn
n=0
∞
X
3 n+1
x
,
=
2n
n=0
−2 < x < 2.
P
n
This time we get the interval of convergence by noting that the equality (1 − x/2) −1 = ∞
n=0 (x/2)
is valid for −1 < x/2 < 1.
The last three power series were found via algebraic manipulation of a known series. Other
power series may be found via integration or differentiation of a known series, as per the following
theorems.
P
n
Theorem 5.4 (Term-by-Term Differentiation) Suppose the series ∞
n=0 cn (x − a) converges
on the interval (a − R, a + R) for some R > 0. On this open interval define the function f to be
23
the series sum:
f (x) :=
∞
X
n=0
cn (x − a)n ,
|x − a| < R.
Then f is differentiable on (a − R, a + R) with expression for f 0 obtained through term-by-term
differentiation of the series:
∞
X
f 0 (x) =
ncn (x − a)n−1 .
n=1
Note that two assertions are being made in this theorem:
(i) The derivative of a power series (an “infinite polynomial”) is obtained via term-by-term
differentiation, just as one expects from polynomials (finitely-long).
(ii) The radius of convergence for the derivative is the same as that of the original series.
By extension, one may get higher derivatives of f by repeating the process. For instance, in the
same interval (a − R, a + R) we have
∞
X
f 00 (x) =
n=2
n(n − 1)cn (x − a)n−2 .
Theorem 5.5 (Term-by-Term Integration) Suppose the series
the interval (a − R, a + R) for some R > 0. Then the series
P∞
n=0 cn (x
− a)n converges on
∞
X
cn
(x − a)n+1
n
+
1
n=0
converges for |x − a| < R. If we define the function f on the domain (a − R, a + R) to be the series
sum
∞
X
f (x) :=
cn (x − a)n ,
|x − a| < R,
n=0
then
Z
f (x) dx = C +
∞
X
cn
(x − a)n+1 ,
n
+
1
n=1
|x − a| < R.
We will use the latter of these theorems to find a power series for arctan x, one of many
P
n 2n
antiderivatives of (1 + x2 )−1 = ∞
n=0 (−1) x . For a specific choice of C, the theorem guarantees
that
∞
X
(−1)n 2n+1
arctan x = C +
x
,
−1 < x < 1.
2n + 1
n=0
We may evaluate C by plugging in x = 0 (the center of the series):
arctan 0 = C + 0 + 0 + · · ·
Thus
arctan x =
∞
X
(−1)n 2n+1
x
,
2n + 1
n=0
24
⇒
C = 0.
−1 < x < 1.
6
6.1
Taylor series
Taylor and MacLaurin series
In the second half of the previous section, we derived formulas for power series of some specific
functions f . Some natural questions arise. First, can a power series be written for any function
f ? Another question comes from the power series of 1/(1 − x), which we know equals the function
only in the interval −1 < x < 1. How does one determine a similar region of equality between an
arbitrary f and its power series? Are they equal at each x value where the power series converges?
How would one go about deciding where a series should be centered, or what the appropriate
coefficients cn should be?
To answer these questions, suppose first that f is equal to its power series representation for all
x in the interval (a − R, a + R). Thus, the series is centered at a and takes the form
∞
X
n=0
cn (x − a)n .
By applying the theorem on term-by-term differentiation repeatedly, it follows that f is differentiable inside (a − R, a + R) up to all orders. Let us evaluate f and its derivatives at x = a. On
the left we find the series expression for various orders of derivatives of f at arbitrary x, and on
the right we draw conclusions about the coefficients c j via evaluating these derivative expressions
at x = a.
∞
X
cn (x − a)n
⇒
c0 = f (a)
f (x) =
n=0
f 0 (x)
=
∞
X
ncn (x − a)n−1
⇒
c1 = f 0 (a)
∞
X
n(n − 1)cn (x − a)n−2
⇒
f 00 (a) = (2 · 1)c2
∞
X
n(n − 1)(n − 2)cn (x − a)n−3
∞
X
n(n − 1) · · · (n − j + 1)cn (x − a)n−j
n=1
f 00 (x)
=
n=2
f 000 (x)
=
n=3
f (j) (x)
=
n=j
⇒
⇒
c2 =
f 00 (a)
2·1
f 000 (a) = (3 · 2 · 1)c3
f 000 (a)
3·2·1
⇒
..
.
c3 =
⇒
f (j) (a) = j!cj
⇒
cj =
f (j) (a)
j!
What we have shown is that, if f equals a power series centered at a in some interval (a − R, a + R),
then the coefficients cn , n = 0, 1, 2, . . ., of the power series representation
f (x) =
∞
X
n=0
cn (x − a)n ,
|x − a| < R
25
are given by
cn =
f (n) (a)
.
n!
That is, if we know already that f has a power series representation in some interval (a − R, a + R)
centered at a, then we may be specific about writing
f (x) =
∞
X
f (n) (a)
n!
n=0
(x − a)n ,
|x − a| < R.
While we still do not have a characterization of which functions f have a power series representation,
a corollary to our work above is that a necessary condition for it to be so is that f be differentiable
of all orders at x = a. While we still do not know whether this is a sufficient condition, it tells us
that, given any f that is differentiable of all orders at x = a, we may write down the only possible
power series representation of f , namely
∞
X
f (n) (a)
n!
n=0
(x − a)n ,
called the Taylor series of f at a. To spread around the honor bestowed upon mathematicians
who helped formulate the ideas of series, the Taylor series of f at 0 is usually called the MacLaurin
series of f . Here are some examples.
1. There is no MacLaurin series of f (x) = |x|, since f is not even differentiable at x = 0, let
alone differentiable up to arbitrary order.
2. One way to find the Taylor series of f (x) = x 2 − 3x + 1 at (−2) is to calculate the coefficients
as stated above:
f 0 (x)
= 2x − 3
⇒
f 00 (x) = 2
⇒
f 000 (x) = 0
⇒
..
.
f (n) (x) = 0
⇒
c0 = f (−2) = 11
c1 = f 0 (−2) = −7
f 00 (−2)
= 1
c2 =
2!
00
f (−2)
= 0
c3 =
3!
f 00 (−2)
= 0,
3!
cn =
for n ≥ 3.
Thus, the Taylor series of f at (−2) is the finite sum
11 − 7(x + 2) + (x + 2)2 .
While, in general, we will not immediately know in these examples whether the given f equals
its Taylor series at a, in this case it is quite easy to expand the result to see that it does give
us f . Such is the case whenever one seeks a Taylor series (“infinite polynomial” expression)
for a polynomial f . One should not be surprised that there were no terms in the Taylor series
beyond the 2nd power of (x − a).
3. Suppose f (x) = ex . Then f (n) (0) = 1 for all n = 0, 1, 2, . . .. This means that the coefficients
cn of the MacLaurin series of f are given by c n = 1/n!, and the full series (at 0) is
∞
X
xn
n=0
26
n!
.
4. Consider the MacLaurin series of another function: f (x) = sin x. A straightforward calculation reveals that


 0, if n is even
(n)
f (0) =
1, if n = 1, 5, 9, . . .


−1, if n = 3, 7, 11, . . . .
Thus, the MacLaurin series of sin x is
∞
X
1
1
1
(−1)n
x2n+1 = x − x3 + x5 − x7 + . . . .
(2n + 1)!
3
5
7
n=0
5. Concerning the function f (x) = (1 − x) −1 we are already a couple of legs up. We already
know a power series representation for this f in the interval −1 < x < 1 which, being centered
at zero, must be the MacLaurin series of f . Moreover, unlike the previous two examples (but
like the 2nd one) we know that, in the interval −1 < x < 1, the two expressions are equal —
that is,
∞
X
1
=
xn ,
−1 < x < 1.
1−x
n=0
Similar statements may be made about
∞
X
1
(−1)n x2n ,
=
1 + x2
n=0
and
6.2
∞
X
(−1)n 2n+1
arctan x =
x
,
2n + 1
n=0
−1 < x < 1,
−1 < x < 1.
A Function to Dash Our Biggest Hopes
Many engineers and researchers in the mathematical sciences assume the equality of a function of
interest and its power/Taylor series, at least in some region local to the center x = a. While often
correct in this assumption, they should keep in mind the function
(
0, if x = 0,
f (x) :=
2
e−1/x , if x 6= 0,
which is differentiable up to all orders at x = 0 and whose MacLaurin series (it may be shown) is
P
n
the zero series (i.e., each cn given by f (n) (0)/n! in the series representation ∞
n=0 cn x turns out
to be zero). Nevertheless, f is not the same as the zero function. This example answers several
questions that have arisen in this section.
• Does every f which is differentiable at x = a up to all orders have a power series
representation at a?
The answer is a bit tricky in its wording. The function will have a Taylor series (itself a power
series) at a but this series need not represent f . That is, there need not be any value besides
x = a at which f and its power series are equal, even when the power series itself has some
nontrivial interval of convergence. (In the example here, the zero series converges for all real
numbers, but the only place it equals f is at x = 0.) Thus, differentiability up to all orders
is a necessary but not sufficient condition for f to have a power series representation.
27
• Will the series at least represent f throughout its interval of convergence?
The answer is ‘no’ (i.e., it need not always be true). See the answer to the previous question
for an explanation.
Those who assume a function f and its Taylor series are equal do so at their own peril.
It can be, in general, an arduous task to show that a given function equals its Taylor series.
We will show it for just a couple of the previous examples. Keep in mind that there were specifics
about Examples 2 and 5 from the last section that tell us, without any additional work, that f
equals the Taylor series. So we will focus our attention upon Examples 3 and 4.
6.3
Taylor’s Theorem, Taylor polynomials and the Equality of a Function and
its Taylor Series
Do you remember the Mean Value Theorem? It says the following:
Theorem 6.1 (Mean Value Theorem) Suppose f is continuous on a closed interval
a ≤ x ≤ b and differentiable on the interval’s interior (a, b). Then there exists a number
c between a and b for which
f (b) − f (a)
= f 0 (c).
b−a
Let us view this theorem from a slightly different perspective. Suppose f is a continuous and
differentiable function in an open interval I containing a. Now suppose x is a number in this
interval (one which is going to take the role of b in the theorem). Then the MVT says that there
is some number c between x and a for which
f (x) − f (a)
= f 0 (c),
x−a
or, rearranging this equation,
f (x) = f (a) + f 0 (c)(x − a).
(8)
This equation tells us something about approximating values f by a linear polynomial at various x
in I away from a, though one must keep in mind that the choice of c (and hence of the coefficient
f 0 (c)) depends upon the choice of x. Notice that the right-hand side of (8) bears a resemblance to
the first two terms of a power series representation of f .
The only assumption necessary to arrive at equation (8) is that f 0 exist in an interval surrounding
a. Can more be said if the derivatives f 00 , f 000 , . . . , f (n) all exist in such an interval? Yes. This is
the content of Taylor’s theorem:
Theorem 6.2 (Taylor’s Theorem) Suppose f is differentiable through order (n +
1) in an open interval I containing a. Then for each x in I, there exists a number c
between x and a such that
f (x) = Tn (x) +
f (n+1) (c)
(x − a)n+1 ,
(n + 1)!
where Tn (x) is the nth-order polynomial defined by
Tn (x) =
n
X
f (j) (a)
j!
j=0
(x − a)j .
Such Tn (x) are called nth-order Taylor polynomials generated by f at x = a.
28
Recall the question we have yet to answer: to determine if the Taylor series of f is, indeed, a
representation of f . Said another way, “Does the Taylor series of f converge to f ?” To answer this
question, we return to the methods of Section 2 and look at the sequence of partial sums of the
series (i.e., the sequence of Taylor polynomials) to see if it converges to f . Taylor’s theorem is the
tool that makes it possible to analyze this convergence.
Example: Convergence of the Maclaurin series generated by e x to ex
What we will show is that, if we fix x at any real number, the sequence of partial sums
x
at that x (i.e., the sequence (Tn (x))∞
n=0 converges to e . We have no worries about
convergence at x = 0, so we assume x 6= 0 (fixed). Since e x has derivatives of all orders,
we are free to choose n to be a nonnegative integer of any size in Taylor’s theorem. Let’s
assume for the moment that n is fixed as well. Focusing on the Taylor series generated
by ex at 0, Taylor’s theorem says that
ex = Tn (x) +
ec
(x)n+1 ,
(n + 1)!
(9)
for some c between 0 and x. If x < 0 then c < 0, which means that e c < 1. On the
other hand, if x > 0 then ec < ex . In either case, ec < e|x| . Subtracting Tn (x) from
both sides of equation (9) and taking absolute values, we get
|ex − Tn (x)| =
≤
ec
|x|n+1
(n + 1)!
e|x|
|x|n+1 .
(n + 1)!
The left-hand side of this equation represents the error in approximating e x by its nthorder Taylor polynomial at a particular x, and the right-hand side provides an upper
bound on that error. Now we let n → ∞. Since
lim
n
|x|n+1
= 0,
(n + 1)!
we have that the error |ex − Tn (x)| → 0 as n → ∞.
A similar analysis can be done to show that, at each real number x,
| sin x − Tn (x)| → 0
as
n→∞
| cos x − Tn (x)| → 0
as
n → ∞,
and
where Tn (x) is the nth-order Taylor polynomial generated by sin x and cos x respectively at 0.
29