Learn Infinite Series some exercises

Chapter 3Infinite Series

SequenceIf to each positive integer n there is assigned a number an, then the numbers an are said to form a sequence. The numbers are often arranged in order, according to subscript as

a1, a2, a3, . . . an

If these numbers are finite then it is a finite sequence otherwise it is said to be infinite sequence.an is called the nth term of the sequence and the sequence is denoted

by {an}. For example , . . . is an infinite

sequence.

Limit of an Infinite SequenceAn infinite sequence {an} is said to converge to the number l or to have a limit l, if for any positive number , there exist a positive number N such that

|an – l| < for all n N

or l - < an < l + for all n N

and we write .

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3Infinite Series


Convergent SequenceAn infinite sequence {an} is said to be convergent if

where l is finite and definite real number.

Divergent SequenceA sequence which diverges to either + or - is said to be a divergent sequence.

Example 1

Show that is a convergent sequence.

Solution

Here

Hence sequence is convergent and its limit is 0.

Monotonic SequencesA sequence is said to be monotonically increasing if an+1 an for all n.A sequence is said to be monotonically decreasing if an+1 an for all n.A sequence is said to be monotonic if it is monotonically increasing or monotonically decreasing.

Bounded SequenceUpper BoundA sequence is said to be upper bound if there exist a number M such that |an| M.

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Lower BoundA sequence is said to be lower bound if there exist a number m such that m |an| .A sequence is said to be bounded if there exists number m and M such that m |an| M, for all n.

Theorem 1Every convergent sequence is bounded.

ProofLet {an} be a sequence which converges to l. Let us choose = 1. Then there exist a positive integer m such that

|an – l| < 1 for all n m

l – 1 < an < l + 1 for all n m

If k = min {a1, a2, . . . , am-1, l – 1}

K = max {a1, a2, . . . , am-1, l + 1}

then k an K for all n

Hence {an} is bounded.

RemarkThe converse of the above theorem does not hold. That is, a bounded sequence may not converge. For example, the sequence {1, -1, 1, -1, . . . . } is bounded, but is not convergent.

Theorem 2A bounded monotonic sequence is convergent.

ProofLet {an} be a bounded nondecreasing sequence i.e.

a1 a2 a3 . . . an . . .Since the sequence is bounded, the set

A = {a1, a2, a3, . . . an, . . . }Whose elements are terms of the sequence, is a bounded set. A1 is a lower bound of A. Since A is also bounded above, it has the supremum. Let M be the supremum of A. Then an M for all n and any number smaller than M is not an upper bound of A. For > 0, M -

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is not an upper bound of A, that is there is some element aN of A such that

M - < aN , N being a positive integer.

Since {an} is a nondecreasing sequence, we have

M - < aN aN+1 aN+2 . . . M < M +

or M - < an < M + for all n N

or |an – M| < for all n N

Hence by definition, an M as n .

Similarly, a bounded nonincreasing sequence {bn} converges and its limit is the infimum of the set

B = {b1, b2, b3, . . . , bn, . . .}

Oscillating SequenceA sequence which is neither increasing nor decreasing is said to be an oscillating sequence. For example 2, -2, 2, -2, . . .

Properties of Sequence

I. If {an} and {bn} are convergent sequences such that an a, bn b and c is a constant, then

(i) c = c

(ii) can = ca

(iii) (an bn) = an bn

Proof

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Let > 0 be given. Since the sequences {an} and {bn} converge to a

and b respectively, therefore there exist positive integers p and q, such

that

|an – a| < /2 whenever n p (1)

|bn – b| < /2 whenever n q (2)

Also since

|an + bn| - (a + b)| = (an – a) + (bn – b)|

|an – a| + |bn – b|

therefore by using (1) and (2), we have

|(an + bn) – (a + b)| < whenever n max{p,q}

Hence {an + bn} converges to a + b

(iv) (anbn) = an bn = ab

(v) , provided b 0.

Proof

Since bn 0 for any n and b 0, therefore, by theorem

(1)

Also

II. For a monotonically increasing sequence a1 a2 a3 . . . , we have

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(i) If the sequence is bounded then an M where M = Sup{a1, a2, ..}.

(ii) If the sequence is not bounded, then an = i.e. sequence is

divergent.

III. For a monotonically decreasing sequence b1 b2 b3 . . . , we have(i) If the sequence is bounded then bn m where m = Inf{b1, b2, …}.

(ii) If the sequence is not bounded, then bn = - i.e. the sequence

is divergent.

Cauchy SequenceA sequence {an} is said to be Cauchy sequence if for given > 0 there exist a positive integer m, such that

|an – am| < where n m

For example is a Cauchy sequence.

Theorem 3Every Cauchy Sequence is bounded.

ProofLet {an} be a Cauchy sequence. Taking = 1, then there must exist a

positive integer m such that

|an – am| < 1 whenever n m

i.e. am – 1 < an< am+1 whenever n m

Let k = min {a1, a2, . . . , am-1, am – 1}

K = max {a1, a2, . . . , am-1, am + 1}

then k an K for all n

Hence {an} is bounded.

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Find for each of the following sequence {an}, whether it converges, diverges or oscillates:

1. 2.

3. 4. nn

5.

The nth term of the sequence is given. Determine whether the sequence converges or diverges. If it converges find its limit:

6. 7.

8. 9.

10. 11.

12. 13.

14. 15. an = ln n – ln (n + 1)

16. 17.

18. 19.

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Exercise 3.1


20. 21.

22. , d> c > 0 23.

24. 25.

Infinite SeriesWe are already familiar with arithmetic and geometric series. In arithmetic series, each term after the first is formed by adding a fixed

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number to the preceding term and in a geometric series each term after the first is multiplying the preceding term by a fixed number. A series can be formed by another ways also. For example, the series 1 + 4 + 9 + 16 + 25 + 36 is formed by squares of the first six natural numbers.A series of n terms is generally denoted by the expression

a1 + a2 + a3 + . . . + an or

Sometimes, a series has no last term, i.e. the terms of the series continue indefinitely. Such a series is called infinite series. An infinite series continue indefinitely by an expression of the form

a1 + a2 + a3 + . . . + an + . . . or

We shall now study some properties of infinite series.

Sequence of partial sums of a series

Let a1 + a2 + a3 + . . . + an + . . . be an infinite series. If Sn denotes the sum of the first n terms of this series, so that

Sn = a1 + a2 + a3 + . . . + an

then {Sn} is called the sequence of the partial sums of the given series. For example, consider the series 1 + 2 + 22 + . . . + 2n-1 + . . . The sum of the first n terms of this series is

Therefore the sequence of partial sums of this series is {Sn}, where Sn = 2n – 1.

Convergent SeriesAn infinite series is said to converge, diverge or oscillate

according as the sequence {Sn} of its partial sums converges, diverges or oscillates.

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If the sequence {Sn} of partial sums of a series converges to S,

then S is said to be the sum of the series .

The following two results are direct consequences of the above definition:

Result 1The alteration or omission of a finite number of terms of a series has no effect on its convergence, divergence or oscillation.

Result 2Multiplication of all the terms of a series by a fixed non-zero real number has no effect on its convergence, divergence or oscillation.

Example 1

Show that the series converges.

Solution

Here

It is a geometric series whose sum can be written as

; where r is the common ratio.

Therefore

As

therefore

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Hence Sn converges to 3. This implies that an converges and

RemarkAs Sn is the sum of n terms. Thus is sum of all terms of the

infinite series, that is

Hence S is the sum of all the terms of the infinite series. If the sum of all the terms is finite then series is convergent.

Example 2The reader is familiar with the geometric series

we investigate the behavior of this series for different values of r.Here

Now

But if |r| < 1 and does not exist if |r| > 1. Thus the

given series

(i) Converges to , if |r| < 1

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(ii) Diverges if |r| > 1

(iii) If |r| = 1, i.e. r = 1, then for r = 1 the series is

a + a + a + . . .

so that the nth partial sum is Sn = na and according as a is

positive or negative. If r = -1 the series isa – a + a – a + . . .

so the sequence of partial sums is a, 0, a, 0, . . . which oscilate. Thus the given series oscilate for r= 1 and r = -1.

Example 3Determine whether the following series converges or diverges

If it converges, find its sum.Solution

The nth partial sum of the series is

Now by using partial fractions we can write,

Therefore

This is a telescopic sum which means that each term cancels part of the next term so that the sum reduces to only two terms. Thus

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Hence series converges and its sum is .

Example 4(The Euler’s Series). Investigate the behavior of the series

+ . . .

Solution

The sequence {Sn} is monotonically increasing, since

S1 = 1, S2 = 1 + ,

We check whether {Sn} is bounded.

< 2 for all n

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Hence |Sn| < 2 for all n and {Sn} is convergent because .

Since the sequence of partial sums of the series is convergent,

is convergent by a theorem.

Cauchy CriterionIf a series is convergent, then for every > 0 there exist a

positive integer N such that|Sm – Sn| < for all m N, n N

Proof.

Let be convergent and its sum be S. Then

|Sm – Sn| = |Sm – S + S – Sn|

|Sm – S| + |Sn – S| (1)

Since for every > 0 there exist positive integers

N1 and N2 such that

and (2)

for all m N1 and n N2 .Let N = max {N1, N2}, then from (1) and the inequality (2), we have

for all m,n N

as required.

Harmonic SeriesA series is said to be harmonic if form an arithematic

series.

Example 5Test for convergence the Harmonic series

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SolutionWe have

if n > 1, then

(1)

if the series were convergent, then by Cauchy Criterion, with ,

m = 2n and n = n, |S2n – Sn| < for n N, where N is sufficiently large. But for (1) it is clear that this inequality is never true and so the series is divergent.

Useful ResultsResult 1.

If converges then .

ProofWe have

so that (1)

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Since the series converges, exists. Let . Since n – 1

as n , we also have . Hence from (1), we obtain

= S – S = 0

Result 2. (A divergence test)

If , then the series diverges.

Result 3.

(i) If and are convergent series with sums S and T, then

the series and are convergent and the sum

and difference of these series are

and

(ii) If converges and diverges then the series

diverges.

(iii) If c is a non zero real number, then the series and

either both converges or both diverges. In case of convergence

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where converges to S.

(iv)The insertion (or deletion) of a finite number of terms into (or from) an infinite series dose not alter its convergence or divergence. However, if the series is convergent then its sum is changed by such insertion (or deletion) . Proof (i), (ii)

Let Sn and Tn denote the nth partial sums of and

respectively. Then

or

therefore

The proof of the remaining parts is similar.

Positive terms SeriesA series is said to be positive term series (or a series of positive

terms) if an > 0 for all positive integers n.

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A Necessary Condition for Convergence.

Theorem 4

A necessary condition for a series to converge is that an 0 as

n .Proof

Let Sn = a1 + a2 + a3 + . . . + an

and Sn-1 = a1 + a2 + a3 + . . . + an-1

Then

an = Sn – Sn-1 (1)

If Sn S as n , then Sn-1 S as n . Now taking limit of both

sides of (1), we obtain

Remarks(i) The students should note that the above condition is only necessary, and not sufficient. There exist non-convergent series for which an 0 as n . For example, consider the series

Here which tends to zero as n tends to infinity, but as proved in

example, the series does not converge.

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(ii)

Example 6

Show that the series does not

converge.Solution

therefore

Since an does not tend to zero as n tends to infinity, therefore, the given series does not converge.

Example 7

Examine the series for convergence.

SolutionHere

therefore

Hence the given series diverges.

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If , then the series diverges.


The basic Comparison TestTheorem 5

Let and be series of the positive terms with an bn for

n = 1, 2, 3, . . . , then

(i) If converges , then also converges.

(ii) If diverges , then also diverges.

Proof(i) Since an bn, for all n, therefore

a1 + a2 + a3 + . . . + an b1 + b2 + b3 + . . . + bn

Let Sn = a1 + a2 + a3 + . . . + an

and Tn = b1 + b2 + b3 + . . . + bn

Therefore

Sn Tn

Given that converges, therefore {Tn} converges and is

finite.

But Sn Tn, therefore

is also finite. This implies {Sn} is convergent.

Therefore is convergent.

(ii)We will prove it by contradiction. To prove diverges we assume

that converges. As an bn for all n, then by (i) must be

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converges which is a contradiction to given condition that is

divergent. Therefore our supposition that converges is wrong.

Hence is divergent.

Example 8

Show that diverges.

Solution

Here

And

i.e.

therefore we choose

hence an > bn for all n.

As diverges therefore also diverges by Basic Comparison

Test.

The limit Comparison Test

Theorem 6

Let and be series of the positive terms,

(i) If = l 0, then either both the series converge or both

diverge.(where l is a finite non zero number)

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(ii) If = 0 and converges then also converges.

(iii) If and diverges then also diverges.

Proof(i)

Given that and are positive terms series and l 0,

therefore l > 0, therefore

By definition for a given > 0 a positive integer N such that

for all n N

for all n N (1)

Right half of inequality (1) gives

for all n N

for n N (2)

Behavior of remains the same as .

From (2) by basic comparison test if converges then is

convergent. And if is divergent then is also divergent.

Left half of inequality (1)

for all n N.

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for n N (3)

Again by basic comparison test if converges then is

convergent. And if is divergent then is also divergent.

(ii)

If

By definition

for all n N

for all n N

for all n N

If converges, converges. Therefore converges by

basic comparison test.(iii)

If

By definition

for all n N

as both series of positive terms therefore,

for all n N

for all n N

If diverges, diverges. Therefore diverges by basic

comparison test.

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Example 9Use the Limit Comparison Test to determine whether the series

converges or diverges.

SolutionThe general term of the series is a quotient containing a highest exponent of 1 in the numerator and a heights exponent of 2 in the denominator. This suggest a comparison with the series whose general term is

therefore

But diverges. Hence given series also diverges.

Integral Test (Cauchy’s Integral Test)Theorem 7

Let be an infinite series of positive terms. If f is a continuous

and decreasing function on [1, [ such that f(n) = an for all positive integers n, then

(i) converges if converges.

(ii) diverges if diverges.

Proof

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f(n) = an

f(n + 1) = an+1

f(x) is decreasing if n < x < n + 1, then

f(n) > f(x) > f(n +1) ; f is a decreasing function.

an > f(x) > an+1

or an+1 < f(x) < an

Integrate from x = n to x = n + 1

Put n =1, 2, 3, . . . , t –1

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. . . . . . . . . .

. . . . . . . . . .

adding the above inequalities, we get

(1)

Take left hand inequality

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If converges, then from calculus is finite and

unique (also by adding a1 on R.H.S. remain finite and unique)

Therefore is finite and unique and also {St} is convergent.

Now take right hand inequality of eq. (1),

< St as it is series of positive terms.

If is divergent then does not exist, therefore

does not exist.

{St} is divergent.

is divergent.

Hence prove the theorem.

Example 10

Use the integral test to show that the Harmonic series

divergence. Solution

Here is a decreasing function , then

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Hence the given series diverges.

Example 11Test for convergence the p-series (Hyperharmonic series)

SolutionFor p > 0, the function f(x) = x-p is positive and decreasing and we can apply the Integral Test. The integral

This limit exists only if the exponent of t is negative, that is , if – p + 1 < 0 or p>1. Thus, the improper integral converges if p> 1 and diverges for 0 < p < 1. If p = 1 the p-series becomes the harmonic series which diverges. Hence the p-series converges for p > 1 and diverges for p 1.

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Determine whether the given series converges or diverges. If it converges, find its sum.

1. 2.

3. 4.

5.

Each of the following is the nth partial sum of an infinite series. Determine the series and check whether it converges.

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Exercise 3.2


6. 7.

8.

9. Prove that if a positive term series converges then the series

converges.

10. Give an example in which and both diverges but

converges.

Using the comparison test, investigate convergence and divergence of the series in Problems (11- 20).

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

In problem 21-40, test each series for convergence or divergence.

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21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34.

35.

36.

37. 38.

39. 40.

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Learn Infinite Series some exercises