1.1
Geometric Series ([Thomas, p763])
Two famous examples of infinite geometric series are
620123 - LECTURE SUMMARY 10
INFINITE SERIES
1+
1 1 1
1
+ + + ··· + n + ··· = 1
2 4 8
2
and
.99999 · · · = 1 .
(†)
Marty Ross
October 15, 2006
This is a summary of my lectures 28-31, roughly equal to lectures 28-30 in the subject
guide. Thomas §11.2-11.6 is an excellent reference for this material.
1
(∗)
Introduction: Definition and Easy Examples
We know that an infinite sequence is simply infinitely many numbers in order:
a1 , a 2 , a 3 , . . . , a n , . . .
We justify the fact that these sums are truly 1 below, but we first note that a general
geometric series is of the form
(!)
S=
∞
!
n=0
arn = a + ar + ar2 + · · · + arn + . . .
1
. (Note that we
So, in (∗) we have a = 12 , r = 21 , and in (†) we have a = .9, r = 10
usually start infinite series with n = 1, but with geometric series it is convention to
start with n = 0: then the first term is ar0 = a).
To make sense of these infinite sums, we first consider the corresponding finite
sums. So, if we sum up to n = N , we are considering
(♣)
SN =
N
!
n=1
We write
{an }
{an }∞
n=1
or
for the whole sequence. An infinite series is where we (try to) sum up the whole
series:
a 1 + a 2 + a3 + · · · + a n + . . .
We use summation notation, and write
!
an
or
∞
!
an
n=1
for the whole series.
Now, it is not at all obvious how to make sense of this, how to add up infinitely
many numbers. In fact, we never really add up infinitely many numbers ! We shall
try to make this clear, by first reviewing geometric series, and then discussing the
general definition. We then consider some further simple examples. However, keep
in mind that there is a sense in which the examples of this Section are misleading:
please note the introduction to §2.
1
arn = a + ar + ar2 + · · · + arN .
There is then a very nice trick for explicitly calculating SN . If we multiply (♣) by
r, we obtain
(♠)
rSN = ar + ar2 + ar3 + · · · + arN +1 .
Then, subtracting (♠) from (♣), almost everything cancels: we are left with
SN − rSN = a − arN +1 .
Solving for SN (assuming r %= 1), we find
"
#
a 1 − rN +1
(♥)
SN =
1−r
Now, the critical point is, to obtain the infinite series, we now let N → ∞:
S = lim SN
n→∞
(as long as this limit exists)
THIS IS WHAT WE MEAN BY THE INFINITE SUM.
2
1.1.1
1.2
Example ([Thomas, p762])
Returning to (∗), since a = r =
SN =
1
2
1
,
2
%
(♥) gives
1−
" 1 #N +1 &
2
1−
1
2
Letting N → ∞, a standard limit gives
=1−
We now
$generalise the above procedure, to define what we mean by a general infinite
series
an . Given an infinite sequence {an }, we define the N ’th partial sum:
1
.
2N +1
'
∞
!
1
1
= S = lim SN = lim 1 − N +1
N →∞
N →∞
2n+1
2
n=0
SN =
(
= 1.
lim SN = S
N →∞
∞
!
n=0
.9 · 10−n = S = lim SN = lim
N →∞
N →∞
'
1−
1
10N +1
(
= 1.
“Convergent” Geometric Series (|r| < 1)
Returning to (!), we see that exactly the same calculation can be performed for any
geometric series, as long as |r| < 1 (so that |r|N +1 → 0). So,
(♦)
S=
∞
!
n=0
arn =
a
1−r
3
|r| < 1 .
∞
!
an = S
n=1
an diverges if {SN } diverges. Similarly, we define
$
an diverging to ±∞.
The second question is obviously at least as difficult as the first (and in general is
much more difficult). We shall spend almost all our time on the first question.
We shall now review geometric series, and give a couple other simple examples.
Again, we emphasise that these examples are somewhat misleading: the trouble is,
these examples are simple enough that we can actually answer the second question
above, in that we can actually calculate S. This will usually not be the case.
1.3
1.1.3
$
⇐⇒
Again, we emphasise, we never really sum up infinitely many numbers: what we
do is sum up the first N numbers, and then let N → ∞. It is $
also very important
to distinguish between two possible questions for a given series
an :
$
$
• Does
an converge? (i.e. does
an make sense as a number S?)
$
$
• What IS
an ? (i.e. what number is
an , what is S?)
1
Returning to (†), since a = .9 and r = 10
, (♥) gives
%
" 1 #N +1 &
.9 1 − 10
1
SN =
= 1 − N +1
1
10
1 − 10
.99999 · · · =
a n = a 1 + a 2 + · · · + aN .
Letting N vary, this gives
$us a new sequence {SN }, the sequence of partial sums.
We then DEFINE that
an converges if {SN } converges, and
Example ([Thomas, p765])
Letting N → ∞, a standard limit gives
N
!
n=1
Also
1.1.2
Definition of an Infinite Series ([Thomas, p763])
Telescoping Series
What we showed in §1.1 is that if |r| < 1 then the geometric series
and
∞
!
a
.
arn =
1
−
r
n=0
$
arn converges,
(In §2.3 we show that geometric series diverge when |r| " 1).
The point was, we were able to obtain an explicit expression for SN , and so our
task was simply to evaluate the limit of the sequence {SN }. We now consider a couple
more examples where SN can be explicitly calculated. Such series are generally called
telescoping.
4
1.3.1
2
Example ([Thomas, p765])
Consider
∞
!
log
n=1
Then
'
n+1
n
(
.
' (
' (
' (
'
(
2
3
4
N +1
+ log
+ log
+ · · · + log
1
2
3
N
'
(
2 3 4
N +1
= log
· · ···
1 2 3
N
SN = log
= log(N + 1) .
"
#
$
diverges to ∞, and we write
So SN → log(∞) = ∞. Thus
log n+1
n
∞
!
log
n=1
1.3.2
'
n+1
n
(
= ∞.
∞
!
n=1
By partial fractions,
$
In all of the above examples, we were able to tell whether an converged or diverged
by explicitly calculating SN , and thus S as well. $
This will generally not be the case:
what we shall develop are TESTS for whether
an converges or diverges, which
don’t depend upon calculating SN . So, in most cases we will not explicitly determine
SN or S.
What this is reminiscent of is the Monotonic Sequence Theorem, which tells us
that certain sequences converge whilst not giving us any clue as to their limits. In
fact, as we shall see, MST is at the heart of almost all of our tests.
2.1
An Illuminating Example
We shall now consider a specific series, and argue that it converges. Afterwards, we
formalise our argument into a test, called the Comparison Test.
The series we shall consider is
∞
!
1
n2n
n=1
Example ([Thomas, p765])
Consider
Tests for Convergence and Divergence
One could imagine playing with the finite sum SN , to try to get an explicit expression,
but it is not at all obvious how. On the other hand, just comparing term by term,
it is obvious that
1
.
n(n + 1)
SN =
1
1
1
= −
.
n(n + 1)
n n+1
1
1
1
1
1
1
1
1
+
+
+ ··· +
# 1 + 2 + 3 + ··· + N
1 · 21 2 · 22 3 · 23
N 2N
2
2
2
2
But the sum on the right is exactly the N ’th partial sum of a convergent geometric
series. The whole sum of that geometric series is 1 (§1.1.1). So, for any N ,
So,
SN =
'
1 1
−
1 2
(
+
'
1 1
−
2 3
Letting N → ∞, we see that
(
+
$
'
1 1
−
3 4
1
n(n+1)
∞
!
n=1
(
+ ··· +
'
1
1
−
N
N +1
converges, with
1
= 1.
n(n + 1)
5
(
1
=1−
N +1
SN =
1
1
1
1
+
+
+ ··· +
#1
1 · 21 2 · 22 3 · 23
N 2N
We don’t particularly care about the 1 on the RHS: the point is that the sequence
{SN } is bounded above. Further, since we’re adding a positive quantity each time,
{SN } is an increasing sequence. So, this is exactly the scenario of$
MST: we conclude
1
that {SN } converges (to God knows what). Then, by definition,
converges!
n2n
6
2.2
The Comparison Test (easy version) [Thomas, p777]
We now formalise the previous argument for general positive series. The result is
called the Comparison Test:
Suppose that 0 # an # bn . Then
$
$
an converges
 bn converges =⇒
(CT)
$
an diverges
=⇒
$
bn diverges
So, in our above, we had an = n21n and bn = 21n ; the proof of the Comparison
Test in general follows from exactly the same argument as we gave in the particular
case. (Note that the second statement logically follows immediately from the first).
It is also good to keep in mind the intuition: we can read the first statement as if
the larger series converges then so does the smaller series; we can read the
second statement as if the smaller series diverges then so does the larger
series. This intuition is much easier to remember than any formulaic version.
It is also important to note, the Comparison Test only works for positive
series (i.e. when each an " 0). As a more pragmatic manner, if we want to
use the Comparison Test, we had better have some series we know about to use
as comparisons! At the moment, we only have geometric series, and a couple of
telescopes. The other series which are very important for comparison are what are
called p-series: see §2.4. So, we’ll give a simple example here, but many more
examples (and other versions of the Comparison Test) will be considered later.
2.2.1
"
#
$
And, from §1.3.1, we know that
log n+1
diverges. So, if we set,
n
'
(
'
(
2n + 7
n+1
bn = log
an = log
n
n
" 2n+7 #
$
then
log n
diverges by the Comparison test.
We make a couple of simple remarks about this example.
• We have used the calculation n + 1 # 2n + 7. This is a bit lazy, and we
could have more precisely written n + 1 < 2n + 7. Still, what we’ve written is
correct (just as it is true that 2 # 3). More to the point, in the context of the
Comparison Test, we simply don’t care whether < is possible, and it’s just a
distraction to worry each time whether to write < or #.
• The argument is simply that since the smaller series diverges, the larger series
diverges as well. Note, however, that to fit in with how we’ve written the
Comparison Test it makes a difference as to which series is called an and which
is called bn . The point is, when given a sequence to test, we don’t automatically
label our series an : we wait to see what other series will be used as a comparison.
Once we have the two series together, then we can see how they fit in with the
Comparison Test.
Example
Consider
∞
!
n=1
Note that
'
(
'
(
2n + 7
n+1
log
# log
n
n
log
'
2n + 7
n
(
.
(since n + 1 # 2n + 7, and log is increasing).
7
8
2.3
The DWMT Test [Thomas, p766]
We shall have many more examples of the Comparison Test, but we first give a
simpler test, a test for divergence. To see what this test says, suppose we have a
convergent series
∞
!
an = S .
n=1
2.3.1
If |r| "
if a %= 0) then arn ! 0. Thus, by the DWMT Test, the geometric
$1 (and
series
arn will diverge.
Note that the DWMT Test includes the possibility that an is divergent, and
the possibility that {an } converges but to something other than 0. Both cases are
illustrated here. For example, taking a = 1 and r = 1, we have
∞
!
If we write out the N ’th and N + 1’st partial sums, then
SN +1 = SN + aN +1
This is just the obvious fact that we obtain a next partial sum by adding one more
term to the preceding partial sum. Rearranging, we have
aN +1 = SN +1 − SN .
$
But SN +1 , SN → S. Now, since
an is assumed convergent, we have S is welldefined and finite. This means we can take limits in (∗), giving
n=0
(DWMT)
If
an ! 0
then
∞
!
an diverges.
n=1
Our name is flippant (it is often called the “divergence test”, and Thomas calls
it the “n’th term test”), but we are making a point here. If a series converges, then
the partial sums SN must be leveling off. The DWMT Test is saying very simply
that (given the hypothesis), this leveling off cannot happen. That is, unlike the other
tests we consider, it is extremely unsubtle.
9
1n =
∞
!
n=1
1 = diverges to ∞.
In this case each an = 1 → 1 (so an ! 0), and SN = N → ∞.
Next, consider taking a = 1 and r = −1. This leads to the series
∞
!
(∗)
aN +1 → 0 − 0 = 0 .
$
It follows that, if the series
an converges, then the sequence {an } converges
to 0. We now turn this logically around, and ask what if {an } does not converge to
zero? This is the Don’t Waste my Time Test:
Example: Divergent Geometric Series (|r| " 1)
n=0
(−1)n = 1 − 1 + 1 − 1 + 1 . . .
In this case an = (−1)n , which diverges: in particular, an ! 0. The N ’th partial
sums in this case are alternatingly 1 and 0, so {Sn } also diverges.
2.3.2
WARNING: DWMT is a one-way test!
DWMT tells us what happens if {an} fails to converge to 0. It is very important to
note that if {an } DOES converge to
"
#DWMT tells
" us
# nothing!
$ 0, then
Consider, for example, the series
log n+1
. Then log n+1
→ log 1 = 0, but
n
n
nonetheless the series diverges (§1.3.1). By comparison, geometric series with |r| < 1
are examples of convergent series with an →
$ 0. Thus, the fact that an → 0 cannot,
in and of itself, tell us whether the series
an converges or diverges: we’ll always
need more information.
$
It’s really a key point of infinite series that if an → 0 then
an may or may not
converge, and determining which can be extremely tricky.
10
2.4
p-Series and The Integral Test
2.4.2
We want to return to the Comparison Test, but we first have to build up our stock
of comparison series. We’ll consider what are called p-series (§2.4.2). In order to
determine the convergence of these series, we’ll need a new test, called the Integral
Test (§2.4.3). Then, in §2.5, we’ll return to the Comparison Test.
2.4.1
The Harmonic Series [Thomas, p772]
∞
!
1
1 1 1
1
= + + + ··· + + ...
n
1 2 3
n
n=1
It is an extremely
important series, both for us, and in mathematics generally. We’ll
$1
is divergent. To do this, consider S1 , S2 , S4 , . . . . We’ll prove that
show that
n
S2 M ≥ 1 +
M
.
2
This is a very slow growth estimate: for example it estimates that the sum of the
first million terms is about 4. Nonetheless, the estimate shows that, no matter how
large L we want the partial sum SN to be, if we choose N = 22L , then SN " L.
To prove (∗), first note that immediately
S1 = 1
1
S2 = 1 +
2
Then, for S22 = S4 , we simply use the fact that 13 " 14 . So,
(
'
1 1 1 1
1 1
1 1
1 1
S4 = + + + " + +
+
=1+ + .
1 2 3 4
1 2
4 4
2 2
"
So
Then for S8 , we use the fact that
'
( '
(
'
( '
(
1 1
1 1
1
1
1
1 1
1
1
3
S8 = + +
+
+
+ ··· +
" 1+ +
+
+
+ ··· +
= 1+ .
1 2
3 4
5
8
2
4 4
8
8
2
1 1 1
, ,
5 6 7
1
.
8
Then to get S16 , we add on 8 new terms, each larger than
another 12 . And so on.
The p-Series are the generalisations of the harmonic series, where we take powers
of n:
∞
!
1
1
1
1
1
= p + p + p + ··· + p + ...
p
n
1
2
3
n
n=1
Thus p = 1 gives The Harmonic Series. As other common examples, we have
 ∞
! 1
1
1
1
1



= 2 + 2 + 2 + ··· + 2 + ...
(p = 2)


1
2
3
n
 n=1 n2
The Harmonic Series is
(∗)
1
,
16
which contributes

∞

!

1
1
1
1
1


√ = √ + √ + √ + ··· + √ + ... .

n
n
1
2
3
n=1
"
p=
1
2
#
The key result is
($)

∞
 convergent
!
1
=
p

n
n=1
divergent
p>1
p#1
How can we prove these results? The convergence for p > 1 we’ll delay to the next
section, but the divergence for p # 1 can be justified now. First of all, the case
p = 1 is just the divergence of the Harmonic Series, which we showed in the previous
section. But if p # 1 then np # n. Thus
1
1
# p
n
n
p # 1.
This
$ 1 means we can apply
$ 1the Comparison Test (with an =
.
diverges, so does
n
np
1
n
and bn =
1
):
np
since
Note that we can also make$the comparison n12 # n1 , but this comparison doesn’t
1
help: showing that our series
is smaller than a divergent
doesn’t prove
n2
$ series
1
converges ...” But
anything.
(That
is
the
Comparison
test in this cases says “If
n
$1
since
doesn’t in fact converge, the Test is simply silent).1
n
1
11
p-series [Thomas, p774]
For a very nice geometric argument that
$
1
n2
12
converges, see Thomas, p 771.
2.5
The Integral Test
Combining these two estimates, and rewriting them in terms of SN , we have
To prove the convergence of the p-series for p > 1, we repeat the Riemann sums
argument used in the analysis of Euler’s Constant (Handout 8, §7.4).2 So, in general,
we’re considering a decreasing sequence an , and we assume we have a decreasing
function f (x) which agrees with the sequence:
.N
f (x) dx # SN # a1 +
.N
f (x) dx
1
1
Taking the limit as N → ∞, we see that the series and the integral converge/diverge
together. That is the Integral Test:
f (n) = an .
Suppose f (x) is a positive and decreasing function and suppose
f (n) = an .
(IT)
Then
∞
$
n=1
Figure 1: Approximating
$
an by
-
a2 + a3 + · · · + aN #
.N
f (x).
f (x) dx .
1
The upper rectangles give us the estimate
(U)
.N
2.5.1
-∞
f (x) dx converges
1
f (x) dx # a1 + a2 + a3 + · · · + aN −1 .
1
2
The argument, in slightly different form, is also the key to Problem 92 in the 123 Problem
Booklet.
Example: Convergent p-series
We now apply the integral test to prove that p-series converge for p > 1. To this
end, we consider the function
1
f (x) = p .
x
Then
.∞
1
13
⇐⇒
Note that the Integral test is a 2-way test: as long as we can evaluate
$(or estimate)
the corresponding integral, then we can definitely determine whether an converges
or diverges.
If we sum the lower rectangles, this will have less area than the area under the
graph of f (x), and we have
(L)
an converges
1
dx = lim
N →∞
xp
.N
1
/
0N
1
dx = lim (1 − p)x(1−p) 1 = p − 1 − (p − 1) lim N (1−p) .
N →∞
N →∞
xp
(1−p)
If p > 1 then N
→ 0, and thus the series converges.
Note that the same calculation shows that the p-series diverge for p < 1, and a
similar argument also works for p = 1 (where the integral gives us log N ).
14
2.5.2
Example
2.5.3
To give another example of the Integral Test, consider the series
∞
!
n=2
1
.
n log n
Before applying the Integral Test, we could first try to use the Comparison Test. In
fact, two obvious attempts are contained in the two inequalities
1
1
1
#
#
(n " 3 for the second inequality)
n2
n log n
n
$ 1
But all this shows is that
is larger than a convergent series (no help) and
n log n
smaller than a divergent series (no help).
1
. Note that the summation
To apply the Integral Test, we define f (x) = x log
x
starts at n = 2, since the corresponding n = 1 term is undefined. So, we’ll apply the
Integral Test with the same lower limit x = 2. We then evaluate
.∞
2
1
dx = lim
N →∞
x log x
.N
2
1
dx = lim
N →∞
x log x
u=log
. N
u=log 2
/
0log N
1
du = lim log u log 2 = ∞.
N →∞
u
Note that we’ve used the substitution u = log x, changed the$limits accordingly, and
1
used the fact that log(log N ) → ∞. The conclusion is that
diverges.
n log n
This is actually a very indicative example: no matter how many series we have
to use as comparison, an evil mathematician can always come up with a new series
which has to be handled from scratch. For example, consider the two series
∞
!
n=2
1
n log n log log n
∞
!
n=2
1
,.
n(log n)2
$ 1
Both are smaller than
, and so that comparison is no help. There is nothing
n log n
else to do but apply the Integral Test (see Problems 108 and 111 in the 123 Booklet)
Interlude: Some Fun Stuff on p-series∗
Exact Evaluation of p-series
Note that we proved convergence of p-series for p > 1, but made no attempt
to try to explicitly evaluate the sums. In fact, if p is even there are nice explicit
expressions. For example
∞
!
1
π2
=
2
n
6
n=1
∞
!
π4
1
=
.
4
n
90
n=1
These are quite striking sums: how on Earth did π get in there?!3 What is perhaps
$ 1
even more striking is if we then consider the odd powers, for example the sum
.
n3
If we ask what this series is, the answer is that no one knows! There is no known
simple expression for this infinite sum!
A Problematic Playtoy
Consider an infinite set of children’s block, where the n’th block has dimensions
1 × n1 × n1 . The volume of the n’th block is n12 . So, the total volume V of the blocks
is
∞
!
π2
1
=
< ∞.
V =
2
n
6
n=1
Now let’s consider the surface area of the blocks. just considering one side 1 × n1 of
the n’th block, we see the surface area of the n’th block is at least n1 . So, the total
surface area A of the blocks is at least
A"
∞
!
1
= ∞.
n
n=1
This is really strange: what is says is that we’d need a finite amount of wood to
make these blocks, but an infinite amount of paint to paint them!
This example (in slightly different form) was first given by the 17th Century
mathematician Evangelista Torricelli, and it caused much consternation at the time.
Indeed, it takes some clear thought to untangle this “paradox”.
3
We’ll actually be able to prove the first of these sums shortly. The second sum, and such sums
in general, are better handled by the techniques of complex analysis. Such techniques are taught in
the subject 620-252.
15
16
2.6
The Comparison Test (limit version) [Thomas, p 778]
We now return to the Comparison Test. We’ll consider some sequences which
“should” converge by Comparison but don’t quite work. We’ll then introduce the
Limit Comparison Test, which is flexible enough to work for these series as well.
2.6.1
Example
Consider
∞
!
n=2
1
.
n2 + 1
We already noticed (§1.3.2) that this is a converging telescoping series. We just note
here that it also converges by comparison, since
1
1
# 2.
n2 + 1
n
2.6.2
Example
Consider
Example
Consider
∞
!
n=1
1
1
# 2
.
n2
n −1
However,
this time the comparison goes in the wrong direction, since it only says
$ 1
is larger than a convergent series.
n2 −1
$ 1
$ 1
There are still easy ways to prove convergence of
. For example,
n2 −1
n2 −1
telescopes the same way as for the previous example. As an alternative proof,4 we
can show
' 2
(
n2
n
n2
n2 − 1 =
+
−1 "
if n " 2.
2
2
2
$ 1
$ 2
Taking reciprocals, we see
converges by comparison with
.
n2 −1
n2
Hereafter known as the Adrian Method.
17
7n − 1
.
2n3 − 3n2 + 5
As with the previous two examples, we think of the terms as being roughly (a factor
times) n12 . However, there is no non-painful way to hammer the comparison: we
fundamentally need a more subtle version of the Comparison Test. We’ll introduce
that Test, and then return to this example.
2.6.4
The Limit Comparison Test
$
$
As we have stated it, to compare the series an and bn , we require that the terms
of the series satisfy
0 # an # b n .
To see how we can weaken this requirement, we first note that it is enough to have
0 # an # Lbn
(∗)
∞
!
1
.
2−1
n
n=2
$ 1
Again, the obvious comparison is with
, and we have
n2
4
2.6.3
for all n " M,
where M and L are any fixed$
positive constants. The point is, the$terms an for
n < M will obviously change
an , but they can’t change whether
an actually
$
$
converges. Similarly, if
bn is a convergent series then so is
Lbn , and so the
inclusion of this factor makes no difference to its use of as a comparison.
What we do is to show how to (in effect) guarantee (∗) in a natural way. We do
this by considering the limit of abnn . This is the Limit Comparison Test.
Suppose that an , bn " 0 (for n " M ), and suppose
lim
n→∞
(LCT)
Then
$
 bn converges
$
an diverges
an
= L < ∞.
bn
=⇒
=⇒
$
$
18
an converges
bn diverges
We will make a number of remarks about this Test, but we first apply it to the
previous example. In this case, we choose
an =
Then
7n − 1
2n3 − 3n2 + 5
7n−1
2n3 −3n2 +5
1
n2
bn =
1
.
n2
7
(7n − 1)n2
→
2n3 − 3n2 + 5
2
$
$ 1
converges, we conclude that
by standard
manipulation. Since
bn =
n2
$
$ limit
7n−1
an =
also converges, by the Limit Comparison Test.
2n3 −3n2 +5
an
=
bn
=
2.6.5
Example
Consider
an # (L + 1)bn
We think of the n’th term as n n12 . Since n beats any power of log n (standard
limit), this suggests the series should be convergent, and we choose
an =
Then
• There are a number of ways to word the Limit Comparison Test, and we have
chosen to word it so that the conclusions read the same as our original Comparison Test. However, our formulation is also artificially restrictive:
first of
$
all, we have
an and which
$ to be careful in choosing which series we label
we label
bn (the previous example wouldn’t have worked if we switched the
labels); secondly, we have ruled out the case L = ∞. Note that L = ∞ implies
that an is eventually much larger than bn : thus we expect the same conclusions,
but with the roles of an and bn reveresed.
• An alternative approach to Limit Comparison is to consider separately the
three cases L = 0, 0 < L < ∞ and L = ∞. Separated this way,
$
$
$
$

L=0
bn conv =⇒
an conv (and
an div ⇒
bn div)



$
$
0<L<∞
bn conv ⇐⇒
an conv


$
$
$
$

L=∞
an conv =⇒
bn conv (and
bn div ⇒
an div)
Note the complete equivalence if 0 < L < ∞, implying we needn’t worry which
sequence is labeled an and which labeled bn . If L = ∞, then the roles of an
and bn are reversed.
19
Thus
for large n.
This is enough to guarantee the comparison conclusions.
n3
n=2
(log n)5
2.6.6
$ (log n)5
n3
(log n)5
n3
bn =
1
.
n2
an
(log n)5
=
→ 0.
bn
n
We consider more examples below, but we now make some important remarks:
• We think of the limit condition as guaranteeing an # bn for large n, but this is
somewhat misleading, particularly if L = 0. What we actually get is
∞
!
(log n)5
converges by Limit Comparison.
Example
Consider
∞
!
(log n)5
n=2
n2
$ 1
This series seems harder, since comparison with
fails (we get L = ∞). However,
n2
$ 1
5
(log n)
√
we can also compare to
→ 0, we again have a convergent series.
3 . Since
n
n2
2.6.7
Example
As a last example, we
" give
# another proof that the Harmonic Series converges, by
comparison with log n+1
. One can show that log(1 + x) # x, but this gives the
n
comparison in the wrong direction. However, taking ratios, L’Hôpital implies
"
#
−1
log n+1
1
n
n2
lim
= lim −1
·
1
1 → 1.
n→∞
n→∞ 2
1
+
n
n
n
"
#
$1
Thus, taking an = n1 and bn = log n+1
, we conclude that
converges.
n
n
20
2.7
The Ratio Test [Thomas, p 782]
$
n+1
if we
For a geometric series $ rn , the ratio of successive terms is r rn = r. Thus, $
an+1
have
a
positive
series
a
with
#
r
for
each
n,
then
we
can compare
an
n
an
$ n
to
r . And, as for the Limit Comparison Test, we need only worry about the
comparison for n large: thus we need only worry about an+1
for n large. This leads
an
immediately to the Ratio Test:
2.7.2
Consider
an+1
= r.
n→∞ an
Then

r<1



r>1



r=1
(RT)
2.7.1
=⇒
=⇒
=⇒
$
$
an converges
an diverges
no conclusion
n
an+1
=
an
Thus
$ 2n
n!
∞
!
n
.
2n
n=1
This series is convergent, since the 2n in the denominator overpowers the n in the
numerator,
but direct comparison to 21n fails. With some sneakiness, we can actually
$ " 2 #n
use
as a comparison, but the ratio test is easier:
3
an+1
=
an
n+1
2n+1
n
2n
=
1 n+1
1
·
→ = r < 1.
2
n
2
21
.
2n+1
(n+1)!
2n
n!
=
2n+1 n!
2
=
→ 0 < 1.
2n (n + 1)!
n+1
converges by the Ratio Test.
These examples highlight a subtle point: if we obtain an+1
→ r < 1 we are
an
not actually performing a limit comparison with rn . What we can do is a limit
comparison with r̄n with r < r̄ < 1: this suffices for our purposes.
2.7.3
Stupid Examples
In both these cases
Consider
n!
Note that it is a basic limit that 2n! → 0, but that doesn’t tell us that the series
converges (i.e. DWMT doesn’t help). However, applying the Ratio Test
Consider the two series
Example
∞
!
2n
n=1
Suppose that each an > 0 and that
lim
Example
∞
!
1
.
n
n=1
∞
!
1
n2
n=1
an+1
→ 1.
an
Thus the Ratio Test gives us no conclusion. In fact, exactly the same happens if we
consider the series
∞
!
n4 .
n=1
Even DWMT works for that one!
Thus, there is a sense in which the Ratio Test is very coarse: polynomials are much
touchier than geometric terms, and the Ratio Test simply can’t see the polynomial
terms. Nonetheless, there are many series where geometric terms are the controlling
factors, and in such cases the ratio Test is extremely useful. We’ll definitely see this
when we consider power series (see Handout 11).
22
2.8
Non-Positive Series: Absolute Convergent Test
Except for the DWMT Test, all of the tests we have considered are fundamentally
for positive series: it’s o.k. if some of the an < 0, but we need an " 0 for n after
some point M . We now consider what to do if this is not the case.
As a rule, positive series are much easier to consider than general series. This
is not surprising, since for positive series we have the Monotonic Sequence Theorem
working for us. So, it’s really just the fact that monotonic sequences are simple: they
either level off to a finite number, or march off to ±∞, but they can’t oscillate.
We only consider two cases. The first are what are called Absolutely Convergent Series, where we can effectively ignore the minus signs. We consider those
here. The second case is of Alternating Series (§2.9), where the terms alternate
between being positive and negative. Both of these cases have test which are easy
to apply: again, the point is that they simply leave many series for which we have
no applicable test.
2.8.2
Examples
ACT immediately tells us that the series (∗) converges. On the other hand, ACT
gives us no information about the series (†). Note that ACT doesn’t tell us that
$ (−1)n+1
diverges, but simply gives us no information. In fact, as we shall see in
n
$
(−1)n+1
does converge.
§2.9,
n
2.8.3
Example: The Ratio Test and Absolute Convergence
$
In general, if a series
an is not absolutely convergent, it may still be convergent:
see §2.9. However, there is an important exception:
If
2.8.1
Absolutely Convergent Series [Thomas, p789]
$
$
A series an is said to be absolutely convergent if |an | converges. For example,
∞
!
(−1)n+1
(∗)
n=1
is absolutely convergent, since
∞
!
(†)
n=1
1
n2
|an | diverges by the Ratio Test then
To illustrate and explain this, consider the series
!
1
1
1
= 2 − 2 + 2 − ...
1
2
3
n2
$
$
Then
is convergent. By comparison
(−1)n+1
1 1 1
= − + − ...
n
1 2 3
$
1
n
is not absolutely convergent, since
diverges.
Note
$ that in neither case have we made any claim about whether the original
series an converges or diverges. But we have a very simple test, the Absolute
Convergence Test, which helps:
|bn+1 |
=
|bn |
(2(n+1))!
((n+1)!)2
(2n)!
(n!)2
=
bn =
!
(−1)n
If
(ACT)
That is, if
$
an is absolutely convergent then
|an | converges then
$
an converges.
23
$
an converges
an diverges as well.
(2n)!
(n!)2
(2n + 2)!n!n!
(2n + 2)(2n + 1)
=
→ 4 = r > 1.
(2n)!(n + 1)!(n + 1)!
(n + 1)(n + 1)
$
Thus,
|bn | diverges by the Ratio Test. But, what’s really happening here, what’s
underlying the Test, is that we’re (roughly) comparing |bn | to 4n . So, we’re actually
also proving that
|bn | → ∞ .
But chucking back in the minus signs, we clearly have
bn ! 0 .
$
$
Thus,
$
bn diverges by the DWMT Test!
24
2.8.4
2.9
Remarks and Proof of ACT
$
• We reiterate$
that ACT is a one-way test: If
an is$not absolutely convergent
- that is, if
|an | diverges - it is still possible that
an converges. Again, in
$ (−1)n+1
the next section we shall prove that
is such an example.
n
• For a positive series, the concept of absolute convergence becomes redundant,
$ 1
is
and can’t help us determine the convergence
For example,
n2
1 1 1of series.
1 2 1 = 12 , and we previously proved
absolutely
convergent,
but
simply
because
n
n
$ 1
converges: ACT hasn’t itself contributed anything.
n2
• Though ACT is very easy to apply, and somewhat intuitive, it$is a little tricky
to prove. The trouble is, if some an < 0 then the sequence
an will not be
monotonic, so we can’t immediately apply MST.
What we can do is let
bn = an + |an | .
Note that
0 # bn # 2|an | .
$
$
$
So, if
an is absolutely convergent (i.e. if
|an | converges) then
bn also
converges, by the Comparison Test. But an = bn − |an |, and so by Algebraic
Limit Laws (noting everything on the RHS is convergent!),
!
!
!
an =
bn −
|an | .
That is,
$
Non-Positive Series: Alternating Series Test
For a non-positive series, if Absolute Covergence fails, determining convergence can
be extremely tricky. We shall consider only one special case, what are called Alternating Series. For such a series, the terms alternate between being positive and
negative. Thus, we can write such series as
!

(−1)n+1 an = a1 − a2 + a3 − . . .



∞
an > 0 .
!


( − 1)n an = −a1 + a2 − a3 + . . .


n=1
Note here, we are explicitly writing out the minus signs. So, for example, the series
(∗) and (†) of the previous section correspond to the choices an = n12 and an = n1 .
2.9.1
Alternating Series Test [Thomas, p787]
We shall give a test for convergence of Alternating Series. To understand this test,
keep in mind an example such as
∞
!
n=1
( − 1)n+1
n+1
.
n
Of course, this series immediately diverges by the DWMT test. It makes clear that
in general, if we want an alternating series to converge, we definitely need an → 0.
This is almost, but not quite enough for the Alternating Series Test:
an converges.
Suppose that
(AST)
Then
∞
!
n=1

an > 0



an → 0



an+1 # an .
( − 1)n+1 an and
∞
!
n=1
( − 1)n an converge
Note the extra condition, that {an } must be a decreasing sequence. We give some
examples illustrating the Test, and then discuss the proof.
25
26
2.9.2
Example
2.9.5
For the series
!
1
,
n2
1
we can apply AST with an = n2 , and conclude that the series converges. However,
it is simpler and more natural to knock of the minus signs, and note that the series is absolutely convergent. Thus the series is also convergent by the Absolute
Convergence Test.
2.9.3
(−1)n
Example
!
1
(−1)n ,
n
we can also apply AST, with an = n1 , to conclude that the series converges. Note
that the Absolute Convergence Test does not apply to this series.
2.9.4
Consider the series
!
(log n)5
.
n2
One can apply AST to this series, but it is painful: it is clear than an → 0 (standard
limits), but we also have to show that {an } is decreasing, at least for n large. We
can do that (see the next example), but it is much easier to take absolute values and
then Compare to 13 . Thus, applying Absolute Convergence and Limit Comparison
n2
together, the series converges (and converges absolutely).
2.9.6
For the series
Example
Example
Consider the series
!
!
(log n)5
(−1)n an =
(−1)n
.
n
$
$1
This series is not convergent, since
an is larger than the divergent series
. It
n
is clear that an → 0, but to show that an is eventually decreasing takes a little work.
To do this, we consider the function
f (x) =
Annoying Example
We give an example to show that the condition that {an } is decreasing really is
necessary. Consider the series
∞
!
n=1
( − 1)n+1 an =
(−1)n
2 1 2 1 2 1 2 1
− + − + − + − + ...
1 1 2 2 3 3 4 4
Clearly the an > 0 and an → 0. But, if we consider the even partial sums, we see
S2N =
1 1 1
1
+ + + ··· + .
1 2 3
N
Thus S2N → ∞, and there’s no way the series can converge.
27
(log x)5
.
x
We then calculate that
x · 5(log x)4 ·
f & (x) =
1
− 1 · (log x)5
(log x)4
x
=
(5 − log x) .
x2
x2
5
Thus, once log x " 5 (i.e. x " e ), f (x) is a decreasing function. This shows
that
$ {ann } is eventually a decreasing sequence, and thus AST applies to prove that
(−1) an converges.
2.9.7
Conditional Convergence
$
We say that a series an is conditionally convergent if it is convergent but not
$
$1
5
and (−1)n (lognn) are conditionally
absolutely convergent. So, for example,
n
convergent. By contrast, a positive series can never be conditionally convergent (since
it is either absolutely convergent or divergent). Note that conditional convergence is
not a test: it is simply a description of what we have already determined.
28
2.9.8
A Difficult Example∗
But now, we can write
S2N = S2N −1 + a2N .
Just to illustrate that our Tests don’t cover everything, consider the three series
! sin n
n2
! sin n
n
!
sin n .
The first series is absolutely convergent, since | sin n| # 1, but the other two are not
that obvious: neither is absolutely convergent, and neither is alternating. In fact,
it’s not too hard to show that the third series diverges. However, the middle series
is very tricky: we leave it as a challenge!
2.9.9
Proof of the Alternating Series Test
As with all our Tests, at its heart the Alternating Series Test involves monotonic
sequences. Here, we consider separately the even and odd partial sums. So, consider
!
(−1)n+1 an = a1 − a2 + a3 − . . .
an > 0.
Then, pairing the terms,
(&)
S2N = (a1 − a2 ) + (a3 − a4 ) + · · · + (a2N −1 − a2N ) .
Now, if we’re assuming
(')
an+1 # an ,
Taking limits, (%) implies that {S2N } and {S2N −1 } have the same limit S, and thus
the whole sequence {SN } must converge to this common limit S.
2.9.10
Error Estimates for Alternating Series
Supposing we have applied the AST, we can also obtain an easy estimate as to how
well SN approximates the whole infinite sum S. By (&), the even partial sums {S2N }
increase to S, and similarly the odd partial sums {S2N −1 } decrease to S. Thus, we
have
S2N # S # S2N −1
But S2N and S2N −1 differ by exactly a2N , which thus bounds the error in these partial
sums. We conclude that for any N
|S − SN | # aN +1
As a simple example, suppose we want to approximate
!
(−1)n
to an accuracy of .001. Then we just need to know when an = n12 falls below .001.
That is equivalent to n2 " 1000, which amounts to n " 32. Thus, summing up the
first 31 terms approximates the wholes infinite sum to an accuracy of .001.
then each pair is nonnegative, and thus {S2N } is an increasing sequence. But we can
also use (') to show {S2N } is bounded above:
(&)
S2N = a1 − (a2 − a3 ) − (a4 − a5 ) − · · · − (a2N −2 − a2N −1 ) − a2N # a1 .
Thus {S2N } converges by MST, and similarly the odd partial sums {S2N −1 } converge.
Does that mean we know the whole sequence {SN } converges?$No, not yet. In fact,
everything we’ve said so far also applies to the divergent series (−1)n+1 . The point
is we haven’t yet used the fact that
(%)
an → 0 .
29
1
n2
30