Mth3101 Advanced Calculus Chapter 3

Advanced calculus MTH 3101
Chapter 3, Infinite series of the sequence

CHAPTER 3
INFINITE SERIES
3.1. OBJECTIVES
(1) Series as Sequence
(2) The concept of Convergent Series
(3) Convergence Tests
(4) Interval of Convergence
(5) Uniform Convergence
Students should be able to test for convergence of a series using the
appropriate tests.
3.2 INTRODUCTION
If we try to add the terms of an sequence {an} we get an expression of the form
a1 + a2 + ... + an + ... (1)
which called an infinite series (or just a series) and is denoted by the symbol
¥

åa
n =1
n or åa n

But does it make sense to talk about the sum of infinitely many terms?
Series is related to sequence and it is also a particular sequence. The partial
sum of a sequence will result another sequence possibly converge to a
particular limit. If that is so, this sum of sequence is an infinite series of the
original sequence.
To determine whether or not a general series (1) has a sum, we consider the
partial sums
S1 = a1
S 2 = a1 + a2
S3 = a1 + a2 + a3
and in general,
S n = a1 + a2 + ... + an ,
these partial sums forms a new sequence, which may or may not have a limit.
If lim S n = S exists (as a finite number), then we call it the sum of the infinite
n ®¥

series åa n

33


¥
Definition 1: Given a series åa
n= 1
n = a1 + a2 + a3 + ... , let Sn denote its nth partial

sum:
n
Sn = å=a k= 1
k a1 + a2 + a3 + ... + an

If the sequence {Sn} is convergent and lim S n = S exists as a real number, then
n ®¥

the series åa n is called convergent and we write
¥

åa
n =1
n = a1 + a2 + a3 + ... = S

The number S is called the sum of the series.
Definition 2: If lim S n = +¥ or lim S n = -¥ , then the series åa n is called
divergent.
¥
Thus, when write åa
n=1
n = S we mean that by adding sufficiently many terms of

the series we can get as close as we like to the number S.
Let the sequence { S n } converge to S, then the series is said to converge. S is
called the sum and is written as
¥
lim S n = å an = S
n ®¥
n =1

S n known as partial sums of this series.
¥
1
Example 1: Find the value of ån =1 n(n + 1)
assume to converge. Write

1 1 1
= -
n(n + 1) n n + 1
and the partial sum of the series is
æ 1ö æ1 1ö æ1 1ö æ1 1 ö 1
S n = ç 1 - ÷ + ç - ÷ + ç - ÷ + ... + ç - = ÷ 1 - .
è 2ø è 2 3ø è3 4ø è n n +1 ø n +1
Then,
¥
1 æ 1 ö
lim S n = å= lim ç1 - = = 1 S
÷
n ®¥
n =1 n(n + 1) n ®¥
è n +1ø
Therefore the series converges.

34


2 n-1
Example 2: Find the value of å 2n
. Note that

1 3 7 2n - 1
Sn = + + + ... + n
2 4 8 2 .
1 1 1 1 n
³ + + + ... + =
2 2 2 2 2
Then lim S n = +¥ , and the series diverge .

å ( -1)
n-1
Example 3: . In this case,

ì1, if n is odd
Sn = í .
î0, if n is even
Then lim S n does not exist.

Example 4: Find å n . Then
n(n + 1)
S n = 1 + 2 + ... + n =
2
This series diverge since lim S n = +¥ .
1 1
Example 5: Show that å3 n
converge to
2
.

æ1ö
Since ( un ) = ç n ÷ is a geometric series we have
è3 ø
æ 1ö
1 1 1 1 ç1 - n ÷ 1æ 1ö
S n = + 2 + ... + n= =è
3 ø
3 3 3 3 1- 1 ç 1 - 3n ÷
2è ø
3
1
and lim S n = .
2
Activity: Find the value of the sum
¥
æ 1ö
å log ç1 + n ÷
n =1 è ø
Hint: Write it as
æ 1ö
log ç1 + ÷ log ( n + 1) - log n
=
è nø

35


3.3 CONVERGENCE OF SERIES
Theorem 1: If the series åu n converge then lim un = 0 .
Proof: Let
S n = u1 + u2 + u3 + ... + un
S n-1 = u1 + u2 + u3 + ... + un-1 .

Then un = Sn - S n-1 .Since åu n converge, then
lim S n = L = lim S n -1 .
n ®¥ n ®¥

Therefore,
lim un = lim ( S n - S n-1 ) = S n - lim Sn -1
lim
= L-L=0
Note: The converse is not true.
1
Contradicting example, the series ån is not converge

æ 1 1 1 1 1 ö
S 2 n = ç 1 + + + ... + + + ... + ÷
è 2 3 n n +1 2n ø
æ 1 1 1ö
S n = ç1 + + + ... + ÷
è 2 3 nø
1 1 1
S 2 n - Sn= + + ... +
n +1 n + 2 2n
1 1 1 æ 1 ö 1
> + + ... + n ç= ÷ =
2n 2n 2n è 2n ø 2
Therefore, { Sn } does not satisfy Cauchy’s condition (there is a theorem in R
saying that a sequence {Sn} is convergent iff it is Cauchy sequence). So {S n }
does not converge.
1
Alternatively, the series ån diverges. It can be shown through the following
method:
1
S2 = 1 +
2
1 1 1 1 1 1 1 1 2
S 22 = 1 + + + > 1+ + + = 1+ + = 1+
2 3 4 2 4 4 2 2 2
1 1 1 1 1 1 æ1 1 1 1ö 3
S 23 = 1 + + + ... + > 1 + + + + ç + + + ÷ 1 + .
=
2 3 8 2 4 4 è8 8 8 8ø 2

36


Inductionally, we obtain
n
S 2n > 1 + .
2
¥
1
Then lim S2n = +¥ , it means ån =
n= 1
¥ . It is Contra-positive to the above

theorem.
Example 6: If lim un ¹ 0 then the series åu n diverge.
n n
1. å n +1 diverge since lim
n +1
=1¹ 0.

2n 2n
2. å 2n - 5 diverge since lim
2n-5
=1 ¹ 0 .

3n - 1 3n - 1 3
3. å 4n - 5 diverge since lim = ¹0
4n + 5 4
Note Students do make mistake in relation with the theorem about
convergence which involves lim un = 0 . Need to assert that when å un
converge then lim un = 0 , but when lim un = 0 , not any conclusion can be made:
un possibly converge and possibly. Be careful with this property! Need to note
that if lim un = 0 , then å un need not converge.
Example 7
1 1
1. ån diverge but lim
n
=0 .

1 1
2. å n log n diverge but lim
n log n
=0 .

1 1
3. ån 2
converge but lim
n2
= 0.

Theorem 2:
Let åu n å v be convergent series, then so are the series å cu
and n n (where
c is a constant), å ( u + v ) and å ( u - v ) and
n n n n

(1) å cu = c å u n n

(2) å ( u ± v ) å u ± å v
= n n n n

(3) å ( au + bv ) a å u + b å v
= n n n n

The proof is simple to perform through the definition of series.

37


Proof: (2) Let åu n = u and åv n = v and
sn = u1 + u2 + ... + un
tn = v1 + v2 + ... + vn
Then
sn + t n = ( u1 + u2 + ... + un ) + ( v1 + v2 + ... + vn )
= ( u1 + v1 ) + ( u2 + v2 ) + ... + ( un + vn )
Therefore,
lim ( sn + tn ) lim sn + lim tn
=
=u+v å=u + å v
n n

lim ( sn + tn )
n ®¥
å (u
= n + vn ) .

Then,
å (u n + vn ) åu + åv
= n n .
Activity: Prove formulae (1) and (3) in the theorem as exercise.
EXAMPLE 8 Show that
1 1 1 1 1
1 + + + + + + ... =2 .
3 6 10 15 21
Solution
Let
1 1 1 1 1 1
S n = 1 + + + + + + ... +
3 6 10 15 21 n( n + 1)
2
Then
1 1 1 1 1 1 1 1
Sn = + + + + + + ... +
2 2 6 12 20 30 42 n(n + 1)
æ 1ö æ1 1ö æ1 1ö æ1 1ö
= ç1 - ÷ + ç - ÷ + ç - ÷ + ç - ÷ +
è 2ø è 2 3ø è3 4ø è 4 5ø
æ 1 1ö æ1 1 ö 1
K+ ç - ÷ + ç - = ÷ 1-
è n -1 n ø è n n + 1 ø n +1
2
Then, S n = 2 - and lim S n = 2 .
n +1

38


Activity
1. Show that the series
1 1 1 1
1+ + + + + ... < 2
2! 3! 4! 5!
2. Is the series
1 1 1 1
+ + + ... +
1+ n 2 + n 3 + n n+n
converge, when n becomes infinite ?

3.4 CONVERGENCE TESTS
1
In the earlier discussion, it is shown that the series ån diverges. But the
1 1
series ån 2
converges (will show later). Hence, we use these series ån
1
(diverge) and ån 2
(converge) as a comparison with other series which we
doesn’t know the type of its convergence.
Note: If åu n is a series with positive terms, then the this series converges or
diverges to infinity (+ ¥).
¥
1
Theorem 2: The p-series ån
n =1
p
is convergent if p>1 and divergent if 0<p£1.

Proof the Theorem 2 is based on integral test which we proof later.
Comparison Test (1)
Let åu n and åv n be the series with the positive terms.

(1) If n åv
is convergent and un £ vn for all n, then åu n is also
convergent.
(2) If åv n is divergent and un ³ vn for all n, then åu n is also divergent.
Comparison Test (2) (Limit Comparison test)
Suppose that åu n and åv n are series with positive terms. If
un
lim = c > 0,
n ®¥ v
n

where c is a finite number and c>0, then either both series åv n and åu n

converge or diverge together.
Note: If c = 0 no conclusion can be made.

39


Proof Comparison Test (1)
(1) Let
n n ¥
sn = å uk , tn = å vk , t = å vk
k=1 k=1 k=1

Since the both series have positive terms, the sequence {sn} and {tn} are
increasing (sn+1=sn+an+1³sn). Also tn®t so tn£t for all n. Since uk£vk we have
sn£tn. Thus sn£t for all n. This means that {sn} is increasing and bounded above
and therefore converges by the monotone sequence theorem. Thus å un
converges.
(2) If åv n is divergent, then tn®¥ (since {tn} is increasing) But uk³vk so
sn³tn. Thus sn®¥. Therefore, åu n diverges.
In using the Comparison Test we must have some known series for the
purpose of comparison. Most of the time we use either a p-series converges
or a geometric series.
Proof Comparison Test (2)
Let m and M be positive numbers such that m<c<M. Because un/vn is close to c for
large n, there is an integer N s.t.
u
m < n < M when n>N,
vn
And so
mvn < un < Mvn , when n>N.
If åv n converges, so does å Mv n . Thus, åu n converges by part (1) of the
comparison test. If åv n diverges, so does å mv n and part (2) of the comparison test
shows that å un diverges.
Proof of Comparison test 2 (another way for the proof comparison test 2:)
vn
Let lim = c > 0 , then for e = 1
n ®¥ u
n

vn
$N > 0, ', - c < 1 when n>N.
un
vn v
or -1 < - c < 1 when n > N, or c - 1 < n < 1 + c when n > N or
un un
(c - 1) un < vn < (1 + c) un when n>N. (1a)

This means that if åu n converges, then åv n converges because of the right
hand side inequality of (1a). If å vn converges then åu n converges, because
of left hand side inequality of (1a).

40


The argument for divergence also can be summarized by the same argument
(and the direct effect of the comparison test 1 (part (2)).
Example 9:
n
1. The series å 2n - 5
diverges.
2

n n 1 1
Solution: Since un = ³ 2= = vn and while the series å diverges
2n - 5 2n2
2n n
and using by part (2) of the comparison test we can conclude that the series
n
å 2n 2 - 5 diverges.
5
2. The series å 2n 2
+1
converges.

Solution: Let
1 5
vn = , un = .
n2 2n 2 + 1
un 5n 2 5 1
Then c = lim = lim
= > 0 while the series å 2 converges by
vn 2n 2 + 1 2 n
5
comparison test we can say that the series å 2 is convergent.
2n + 1
n2
3. The series å n3 + 2 diverges.
1 u n3 1
Solution: Let vn =
n
and c = lim = n lim 3
=
n ®¥ v n +2
1 > 0 . The series ån
n
diverges and by limit comparison test we arrive the results.
1
4. The series å ln n diverges.

1 1
Solution: Since ln n < n for positive n, then > . Using part (1) of by limit
ln n n
1
comparison test 1, we can conclude that The series å diverges.
ln n
1
EXAMPLE 5 Test the series å2 n
-1
for convergence or divergence.

SOLUTION We use the Limit Comparison Test with

41


and obtain

1
Since this limit exists and å2 n
is a convergent geometric series, the given
series converges by the Limit Comparison Test.
¥
5
EXAMPLE 6 Determine whether the series å 2n
n= 1
2
+ 4n + 3
converges or

diverges.
SOLUTION: For large n the dominant term in the denominator is 2n2, so we
¥
5
compare the given series with the series å 2 . Observe that
n =1 2n

because the left side has a bigger denominator. (In the notation of the
Comparison Test, is the left side and is the right side.) We know that

is convergent (p series with p=2). Therefore is convergent by part (1) of the
Comparison Test. Although the condition or in the Comparison Test is given
for all n, we need verify only that it holds for n³N, where is some xed integer,
because the convergence of a series is not affected by a nite number of
terms. This is illustrated in the next example.
Problems and Solutions
1
1. Is the series å n2
converge?
æ 1ö
ç1 + ÷
è nø
Solution: Note that
n2 n
æ 1ö éæ 1 ö n ù
ç1 + ÷ êç 1 + ÷ ú > 2
= n

è nø êè n ø ú
ë û
Hence,
1 1
n2
< .
æ 1ö 2n
ç1 + ÷
è nø

42


1
The series å2 n
converges (geometric seies), then the above series also
converges (comparison test).
n+2
2. Is the series å (n + 1)
n+3
converges?

Solution: Note that
n+2 n +1 1 1
> = > , n > 4.
(n + 1) n + 3 (n + 1) n + 3 n +3 n
Then the series diverges by part (2) of comparison test 1.
Activity: The students can try to prove by the same way.

3.5 THE INTEGRAL TESTS; ESTIMATING SUMS
In this section we deal only with series with positive terms, so the partial sums
are increasing. In view of the Monotonic Sequence Theorem, to decide
whether a series is convergent or divergent, we need to determine whether the
partial sums are bounded or not.
Testing with an Integral
Let’s investigate the series whose terms are the reciprocals of the squares of
the positive integers:

There’s no simple formula for the sum of the rst terms, but the computer
generated table of values given in the margin suggests that the partial sums
are approachinga number near 1.64 as and so it looks as if the series is
convergent.
We can conrm this impression with a geometric argument. Figure 1 shows
the curve y=1/x2 and rectangles that lie below the curve. The base of each
rectangle is an interval of length 1; the height is equal to the value of the
function y=1/x2 at the right endpoint of the interval. So the sum of the areas of
the rectangles is

43


If we exclude the rst rectangle,the total area of the remaining rectangles is
smaller than the area under the curve y=1/x2 for x³1, which is the value of the
¥
dx
integral òx
1
2
. We know that this improper integral is convergent and has value

1. So the picture shows that all the partial sums are less than

Thus, the partial sums are bounded and the series converges. The sum of the
series (thelimit of the partial sums) is also less than 2:

[The exact sum of this series was found by the Swiss mathematician Leonhard
Euler (1707–1783) to be p2/6, but the proof of this fact is beyond the scope of
this book.]
Now let’s look at the series

The table of sn values of suggests that the partial sums aren’t approaching a
nite number, so we suspect that the given series may be divergent. Again we
1
use a picture for conrmation. Figure 2 shows the curve y = , but this time
x
we use rectangles whose tops lie above the curve.

44


The base of each rectangle is an interval of length 1. The height is equal to the
1
value of the function y = at the left endpoint of the interval. So the sum of
x
the areas of all the rectangles is

1
This total area is greater than the area under the curve y = for x³1, which
x
¥
dx
is equal to the integral ò
1 x
. But we know that this improper integral is

divergent. In other words, the area under the curve is innite. So the sum of
the series must be innite,that is,the series is divergent.
The same sort of geometric reasoning that we used for these two series can
be used to prove the following test.

NOTE: When we use the Integral Test it is not necessary to start the series or
the integral at n=1. For instance, in testing the series

Also, it is not necessary that be always decreasing. What is important is that
be ultimately decreasing, that is, decreasing for x larger than some number N.
¥ ¥
Then å an is convergent, so
n= N
åa
n =1
n is convergent.

Integral Test
Let f be a monotonically decreasing real positive real function, with the
property f (n) = un , n = 1, 2, 3, ….
¥ ¥
If ò 1
f ( x) dx converges, then the series åu n converges. If ò 1
f ( x) dx
diverges, then the series åu n diverges.
¥ n
(The integral ò 1
f ( x) dx means lim ò f ( x) dx )
n ®¥ 1

45


Proof
Let f(n) = un , n = 1, 2, 3, … be a monotonically decreasing function.
Then, for r an integer
ur +1 £ f ( x ) £ ur , r £ x £ r +1 .
Integrate each term in the above inequality for the interval [r, r + 1], then we
obtain
r +1
ur +1 £ ò f ( x) dx £ ur .
r

Then
n n +1 n r +1 n

å ur +1 =
r=1
å ur £ å
n 2 r 1 =
ò r
f ( x) dx £ å ur .
r 1 = =

Or,
n +1 n +1 n

åu £ ò
n=2
r 1
f ( x ) dx £ å ur .
r =1

Let S n denotes the series partial sum un , that is, S n = u1 + u2 + ... + un , then the
above inequality gives
n +1
S n+1 - u1 £ ò f ( x) dx £ Sn , n = 1, 2, 3...
1

From the above final inequality, note that
lim n +1
(1) If
n®¥ ò 1
f ( x) dx converges, then S n convergesmenumpu .

lim n +1
(2) If
n®¥ ò n
f ( x) dx diverges, then

S n diverges .

46

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