Math 113 Exam 3
November 10, 2016
Exam 3 covers 11.1-11.9 in the textbook.
11.1 Sequences
A sequence is an ordered list of numbers. This can be expressed as {an }
or as a list a1 , a2 , .... We are primarily interested in limits of sequences.
Theorem 0.1 If limx→∞ f (x) = L and f (n) = an , then limn→∞ an = L.
The above says that if we can express the terms of the sequence as the
values of a function where we can compute the limit of the function, then
the sequence has the same limit.
We also have the following properties similar to limits of functions.
Theorem 0.2 If {an } and {bn } are convergent sequences and c is a constant,
then
• limn→∞ an + bn = limn→∞ an + limn→inf ty bn ,
• limn→∞ an − bn = limn→∞ an − limn→inf ty bn ,
• limn→∞ can = c limn→∞ an ,
• limn→∞ an bn = limn→∞ an limn→∞ bn ,
• limn→∞
an
bn
=
limn→∞ an
limn→∞ bn
if limn→∞ bn 6= 0, and
• limn→∞ apn = [limn→∞ an ]p if p > 0 and an > 0 for each n.
1
We also have a squeeze limit theorem that can be helpful with certain
sequences involving trigonometric functions.
Theorem 0.3 If an ≤ bn ≤ cn for n ≥ n0 and limn→∞ an = limn→∞ cn = L,
then limn→∞ bn = l.
Another useful fact is the following.
Theorem 0.4 If limn→∞ |an | = 0, then limn→∞ an = 0.
For continuous functions we also have the following.
Theorem 0.5 If limn→inf ty an = L and f is continuous at L, then limn→∞ f (an ) =
f (L).
For certain sequences that we use specifically related to series we have
the following notions.
A sequence {an } is bounded above if there is some constant M such that
an ≤ M for all n. Similarly, a sequence {an } is bounded below if there exists
some constant m such that m ≤ an for all n. A sequence is increasing if
an+1 ≥ an for each n, and a sequence is decreasing if an+1 ≤ an for each n.
A sequence is monotonic if it is increasing or decreasing.
Theorem 0.6 (Monotonic sequence theorem) Every bounded monotonic sequence is convergent.
11.2 Series
P
A series is an infinite sum of terms of a sequence and is denoted ∞
n=1 an .
Note the initial index may change depending on the situation (for instance
n
0 or n = 2). The n-th partial sum of a series is sn = a1 + · · · + an =
P=
n
i=1 ai . A series is convergent if limn→∞ sn exists and the limit of the partial
sums is the value of the series. If the sequence of partial sums are divergent,
then the series is divergent.P
n)
n−1
and so the geoFor a geometric series ∞
we have sn = a(1−r
n=1 ar
1−r
a
metric series converges to the sum 1−r
if |r| < 1 and diverges if |r| ≥ 1.
The first test you should do for convergence of a series is the following.
Theorem 0.7 If
P∞
n=1
an is convergent, then limn→∞ an = 0.
2
The
following implies that if limn→∞ an does not exist or limn→∞ an 6= 0,
P∞
then n=1 an is divergent.
P
P
Theorem 0.8 P
If
anPand
bn are convergent
series and c is a constant,
P
then the series
can , (an + bn ), and (an − bn ) are convergent. Furthermore, we have the following.
P
P
•
can = c an ,
P
P
P
•
(an + bn ) = an + bn , and
P
P
P
•
(an − bn ) = an − bn .
11.3 Integral test and Estimation of Sums
This sections covers special classes of sequences where f (n) = an for a
function f .
Theorem 0.9 (Integral test)PIf f is continuous, positive, and decreasing
on
R∞
[1, ∞) and f (n) = an , then
an is convergent if and only if 1 f (x)dx is
convergent.
P
c
From this we deduce the following fact. The series ∞
n=1 np is convergent
if p > 1 and divergent if p ≤ 1.
These are one of the classes P
of sequences where we can estimate the
remainder. As a reminder Rn = ∞
i=n ai and is the difference between the
value of a convergent series and the value of the n-th partial sum of the series.
Suppose that f (k) = ak where f is continuous,
R ∞ positive, and decreasing.
P
If
ak is convergent that the remainder Rn ≤ n f (x)dx.
11.4 Comparison Tests
This section covers the comparison test and limit comparison test.
P
P
Test 0.10 (Comparison Test) Suppose that
an and
bn are series with
positive terms.
P
1. If
bn is convergent and an ≤P
bn for all n (or for all n ≥ k where k
is some natural number), then
an is also convergent.
3
P
2. If
bn is divergent and an ≥Pbn for all n (or for all n ≥ k where k is
some natural number), then
an is also divergent.
The advantage to this test is that it can be very straightforward. However,
it can only be used in specific instances. In comparing the two types of series
are
X a
X
and
arn .
np
Remark 0.11 Be careful that you can set up the appropriate inequality. For
instance
1
1
1
1
≤ n but n
≥ n.
n
2 +1
2
2 −1
2
So in the first case use a comparison test, but the easiest thing for the second
case is to use the limit comparison test.
P
P
Test 0.12 (Limit Comparison Test) Suppose that
an and
bn are series
with positive terms. If
an
=c
lim
n→∞ bn
where c is a real number and c > 0, then either both series converge and both
diverge.
This test can be used in most instances when the comparison test can be
used. Example 3 and 4 in the text show good examples of when to apply
this test.
For a straight comparison test we can get estimates on the sum. If
an ≤ bn , Rn = an+1 + an+2 + · · · and Tn = bn+1 + bn+2 + · · · , then Rn ≤ Tn
and Tn can be found sometimes by using the integral remainder in section
11.3. (See Example 5 in the section).
11.5 Alternating Series
Test 0.13 (Alternating Series Test) If the series
satisfies
P∞
n=1 (−1)
n−1
bn (or
P∞
n=1 (−1)
1. bn+1 ≤ bn for all n (or for all n ≥ k where k is some natural number)
2. limn→∞ bn = 0
4
n
bn
then the series is convergent.
This is the easiest test since in general one only needs to show that the
sequence bn is decreasing and convergent. This test is especially helpful for
checking the interval of convergence for a power series since one endpoint is
usually given by an alternating series.
There is also a nice estimate for the remainder given by the following.
P
Theorem 0.14 If s = (−1)n−1 bn and Rn = s − sn where bn is decreasing
and limn→∞ bn = 0, then |Rn | ≤ bn+1 .
See Example 4 in the section for how to apply this.
11.6 Absolute Convergence and the Ration and Root Tests
P
P
Definition 0.15 A series
an is absolutely convergent
if
|an | isPconverP
gent. A series is conditionally convergent if
an converges, but
|an | is
divergent.
Theorem 0.16 If
P
|an | converges, then
P
an converges.
Example 3 in the text gives a good use of this theorem combined with a
simple comparison test.
Test 0.17 (Ratio Test)
an+1 P
= L < 1, then the series ∞ an absolutely converges.
1. If lim n=1
n→∞
an an+1 = L > 1, then the series P∞ an diverges.
2. If lim n=1
n→∞
an an+1 = L = 1, then the test is inconclusive.
3. If lim n→∞
an This test is especially useful for series that involve functions involving
factorials and exponents. Unlike the previous two sections there are no estimates on the convergence.
5
Test 0.18 (Root Test)
1. If lim
n→∞
gent.
p
P
n
|an | = L < 1, then the series ∞
n=1 an is absolutely conver-
2. If lim
p
P
n
|an | = L > 1, then the series ∞
n=1 an is divergent.
3. If lim
p
n
|an | = L = 1, then the test is inconclusive.
n→∞
n→∞
This test is useful when there series is either entirely or mostly terms
involving powers of n.
11.7 Strategy for Testing Series
You should know the chart on page 739 of the text. The steps 1-8 are
helpful in determining which test to use when. Remember though that in
many cases two or three test can be useful.
I would recommend going through most of the odd problems in this section to see that you know how to check for convergence or divergence of series.
11.8 Power Series
Definition 0.19 A power series is a series of the form
∞
X
cn (x − a)n
n=k
where k ≥ 0.
In this first section the main thing we learned was to use the ratio test
to find the radius of convergence for a power series. We had the following
theorem.
Theorem 0.20 For a given power series
∞
X
n=0
possibilities:
6
cn (x − a)n there are only three
1. the series converges only when x = a,
2. the series converges for all x, or
3. there is a positive number R such that the series converges if |x−a| < R
and diverges if |x − a| > R.
The number R is called the radius of convergence. In the third case
above we also check x = a − R and x = a + R in the power series to check
convergence or divergence. To check these endpoints use any of the tests
from the previous sections. There are four possibilities for the interval of
convergence it is (a−R, a+R), [a−R, a+R), (a−R, a+R], or [a−R, a+R].
The examples in the text show how to find the radius and interval for
different examples. In addition the odd problems from 3-27 could be helpful
to give you practice finding the interval and radius of convergence.
11.9 Representations of Functions as Power Series
∞
X
1
xn to find ways
=
1 − x n=0
to express power series for various functions. Besides straight forward substitutions the other technique was the following theorem.
P
n
Theorem 0.21 If the power series
cn (x − a)
of convergence
P has radius
n
R > 0, then the function f defined by f (x) =
cn (x − a) is differentiable
and integrable on (a − R, a + R) and
In this section we mostly used variations on
0
1. f (x) =
∞
X
ncn (x − a)n−1
n=1
Z
2.
f (x) dx = C +
∞
X
n=0
cn
(x − a)n+1
.
n+1
Furthermore, the radius of convergence is the same for f 0 (x) and
and the integral can be definite instead of indefinite.
R
f (x) dx
Using this we found power series expressions for functions like tan−1 x
and ln(x + 1) among others. We were also able to show that f (x) = f 0 (x)
∞
X
xn
for
. So f (x) = ex .
n!
n=0
7
This can be used to find the value of integrals as in example 8 in the
section. We can also use this to find limits as in the following.
tan−1 (x2 ) − x2
.
x→0
x6
lim
8