Introduction to Infinite Series

SEQUENCES AND SERIES

Section 3.7 Introduction to Infinite Series

8. Investigate the convergence or the divergence of the following sequences:

(a) ð ffiffiffiffiffiffiffiffiffiffiffiffiffi n²þ2

p Þ; (b) ð ffiffiffi

pn

=ðn²þ1ÞÞ;

(c) ð ffiffiffiffiffiffiffiffiffiffiffiffiffi n²þ1 p = ffiffiffi

pn

Þ; (d) ðsin ffiffiffi

pn Þ:

9. LetðxnÞandðy_nÞbe sequences of positive numbers such that limðxn=y_nÞ ¼ þ1, (a) Show that if limðy_nÞ ¼ þ1, then limðxnÞ ¼ þ1.

(b) Show that if (xn) is bounded, then limðy_nÞ ¼0.

10. Show that if limðan=nÞ ¼L, whereL>0, then limðanÞ ¼ þ1.

Just as a sequence may be indexed such that its first element is notx1, but isx0, orx5or x99, we will denote the series having these numbers as their first element by the symbols

X¹

n¼0

x_n or X¹

n¼5

x_n or X¹

n¼99

x_n:

It should be noted that when the first term in the series isxN, then the first partial sum is denoted by sN.

Warning The reader should guard against confusing the words ‘‘sequence’’ and ‘‘series.’’

In nonmathematical language, these words are interchangeable; however, in mathematics, these words are not synonyms. Indeed, a series is a sequenceS¼(sk) obtained from a given sequenceX¼(x_n) according to the special procedure given in Definition 3.7.1.

3.7.2 Examples (a) Consider the sequenceX:¼ ðrⁿÞ¹_n¼0wherer2R, which generates the geometric series:

(3) X¹

n¼0

rⁿ¼1þrþr²þ þrⁿþ :

We will show that if jrj<1, then this series converges to 1=ð1rÞ. (See also Example 1.2.4(f).) Indeed, ifsn:¼1þrþr²þ þrⁿforn0, and if we multiplysn

byrand subtract the result fromsn, we obtain (after some simplification):

s_nð1rÞ ¼1r^nþ1: Therefore, we have

s_n 1

1r¼ r^nþ1 1r; from which it follows that

s_n 1 1r

jrj^nþ1 j1rj:

Sincejrj^nþ1!0 whenjrj<1, it follows that the geometric series converges to 1=ð1rÞ when jrj<1.

(b) Consider the series generated by

ð1Þ^{n 1}_n¼0; that is, the series:

(4) X¹

n¼0

ð1Þⁿ ¼ ðþ1Þ þ ð1Þ þ ðþ1Þ þ ð1Þ þ :

It is easily seen (by Mathematical Induction) thats_n¼1 ifn0 is even ands_n¼0 if nis odd; therefore, the sequence of partial sums isð1;0;1;0;. . .Þ. Since this sequence is not convergent, the series (4) is divergent.

(c) Consider the series

(5) X¹

n¼1

1

nðnþ1Þ¼ 1 12þ 1

23þ 1

34þ :

3.7 INTRODUCTION TO INFINITE SERIES 95

By a stroke of insight, we note that 1

kðkþ1Þ¼1

k 1

kþ1:

Hence, on adding these terms fromk¼1 tok¼nand noting the telescoping that takes place, we obtain

sn ¼1

1 1

nþ1;

whence it follows that sn!1. Therefore the series (5) converges to 1. &

We now present a very useful and simplenecessarycondition for the convergence of a series. It is far from being sufficient, however.

3.7.3 The nth Term Test If the seriesP

xn converges, thenlimðxnÞ ¼0.

Proof. By Definition 3.7.1, the convergence ofP

x_n requires that limðs_kÞexists. Since x_n ¼s_ns_n1, then limðx_nÞ ¼limðs_nÞ limðs_n1Þ ¼0. Q.E.D.

Since the following Cauchy Criterion is precisely a reformulation of Theorem 3.5.5, we will omit its proof.

3.7.4 Cauchy Criterion for Series The seriesP

xn converges if and only if for every e>0there exists MðeÞ 2N such that if m>nMðeÞ,then

(6) js_ms_nj ¼ jx_nþ1þx_nþ2þ þx_mj<e:

The next result, although limited in scope, is of great importance and utility.

3.7.5 Theorem Let(xn)be a sequence of nonnegative real numbers. Then the seriesP xn

converges if and only if the sequence S¼ ðskÞof partial sums is bounded. In this case, X¹

n¼1

x_n¼limðs_kÞ ¼supfs_k:k2Ng:

Proof. Sincex_n0, the sequenceSof partial sums is monotone increasing:

s1s2 s_k :

By the Monotone Convergence Theorem 3.3.2, the sequenceS¼ ðskÞ converges if and only if it is bounded, in which case its limit equals supfskg. Q.E.D.

3.7.6 Examples (a) The geometric series (3) diverges if jrj 1.

This follows from the fact that the termsrⁿdo not approach 0 whenjrj 1.

(b) The harmonic seriesX¹

n¼1

1

n diverges.

Since the terms 1=n!0, we cannot use the nth Term Test 3.7.3 to establish this divergence. However, it was seen in Examples 3.3.3(b) and 3.5.6(c) that the sequence (sn) of partial sums is not bounded. Therefore, it follows from Theorem 3.7.5 that the harmonic series is divergent. This series is famous for the very slow growth of its partial sums (see the

discussion in Example 3.3.3(b)) and also for the variety of proofs of its divergence. Here is a proof by contradiction. If we assume the series converges to the numberS, then we have

S ¼ 1þ1 2

þ 1 3þ1

4

þ 1 5þ1

6

þ þ 1 2n1þ 1

2n

þ

> 1

2þ1 2

þ 1 4þ1

4

þ 1 6þ1

6

þ þ 1 2nþ 1

2n

þ

¼1þ1 2þ1

3þ þ1 nþ

¼S:

The contradiction S >S shows the assumption of convergence must be false and the harmonic series must diverge.

Note The harmonic series receives its musical name from the fact that the wavelengths of the overtones of a vibrating string are 1=2, 1=3, 1=4, . . . , of the string’s fundamental wavelength.

(c) The 2-seriesX¹

n¼1

1

n² is convergent.

Since the partial sums are monotone, it suffices (why?) to show that some subsequence of (sk) is bounded. If k1:¼2¹1¼1, thensk1 ¼1. If k2:¼2²1¼3, then

s_k2¼1 1þ 1

2²þ 1 3²

<1þ 2

2² ¼1þ1 2; and ifk3:¼2³1¼7, then we have

sk3¼sk2þ 1 4²þ 1

5²þ 1 6²þ 1

7²

<sk2þ 4

4²<1þ1 2þ 1

2²: By Mathematical Induction, we find that ifk_j :¼2^j1, then

0<skj <1þ¹₂þ ¹₂ 2

þ þ ¹_{2 j}1

:

Since the term on the right is a partial sum of a geometric series withr¼¹₂, it is dominated by 1=1¹₂

¼2, and Theorem 3.7.5 implies that the 2-series converges.

(d) The p-seriesX¹

n¼1

1

n^p converges whenp >1.

Since the argument is very similar to the special case considered in part (c), we will leave some of the details to the reader. As before, ifk1 :¼2¹1¼1, then sk1¼1. If k2:¼2²1¼3, then since 2^p<3^p, we have

sk2 ¼ 1 1^pþ 1

2^pþ1 3^p

<1þ 2

2^p¼1þ 1 2^p1: Further, ifk3:¼2³1, then (how?) it is seen that

sk3 <sk2þ 4

4^p<1þ 1 2^p1þ 1

4^p1:

Finally, we let r:¼1=2^p1; since p > 1, we have 0<r<1. Using Mathematical Induction, we show that ifk_j:¼2^j1, then

0<s_kj <1þrþr²þ þr^j1< 1 1r:

Therefore, Theorem 3.7.5 implies that the p-series converges whenp>1.

3.7 INTRODUCTION TO INFINITE SERIES 97

(e) Thep-seriesX¹

n¼1

1

n^p diverges when 0<p1.

We will use the elementary inequalityn^pnwhenn2Nand 0<p1. It follows that 1

n 1

n^p for n2N:

Since the partial sums of the harmonic series are not bounded, this inequality shows that the partial sums of thep-series are not bounded when 0<p1. Hence thep-series diverges for these values of p.

(f) The alternating harmonic series, given by

(7) X¹

n¼1

ð1Þ^nþ1

n ¼1

11 2þ1

3 þð1Þ^nþ1 n þ is convergent.

The reader should compare this series with the harmonic series in (b), which is divergent. Thus, the subtraction of some of the terms in (7) is essential if this series is to converge. Since we have

s2n¼ 1 11

2

þ 1 31

4

þ þ 1 2n1 1

2n

;

it is clear that the ‘‘even’’ subsequence ðs2nÞis increasing. Similarly, the ‘‘odd’’ subse- quence ðs2nþ1Þis decreasing since

s2nþ1¼1

1 1

21 3

1 41

5

1 2n 1

2nþ1

:

Since 0<s2n<s2nþ1=ð2nþ1Þ ¼s2nþ11, both of these subsequences are bounded below by 0 and above by 1. Therefore they are both convergent and to the same value. Thus the sequence (sn) of partial sums converges, proving that the alternating harmonic series (7) converges. (It is far from obvious that the limit of this series is equal to ln 2.) &

Comparison Tests

Our first test shows that if the terms of a nonnegative series are dominated by the corresponding terms of aconvergent series, then the first series is convergent.

3.7.7 Comparison Test Let X:¼(xn)and Y:¼(yn)be real sequences and suppose that for some K2N we have

(8) 0x_ny_n for nK:

(a) Then the convergence of P

y_n implies the convergence ofP x_n. (b) The divergence ofP

x_n implies the divergence ofP y_n. Proof. (a) Suppose thatP

y_n converges and, givene>0, letMðeÞ 2Nbe such that if m>nMðeÞ, then

y_nþ1þ þy_m<e:

Ifm>supfK; MðeÞg, then it follows that

0x_nþ1þ þx_my_nþ1þ þy_m<e;

from which the convergence ofP

x_n follows.

(b) This statement is the contrapositive of (a). Q.E.D.

Since it is sometimes difficult to establish the inequalities (8), the next result is frequently very useful.

3.7.8 Limit Comparison Test Suppose that X:¼ ðxnÞ and Y :¼ ðy_nÞ are strictly positive sequences and suppose that the following limit exists in R:

(9) r:¼lim xn

y_n : (a) If r6¼0 thenP

xn is convergent if and only ifP

y_n is convergent.

(b) If r¼0and ifP

y_n is convergent, thenP

xn is convergent.

Proof. (a) It follows from (9) and Exercise 3.1.18 that there existsK2Nsuch that¹₂r xn=y_n2rfornK, whence

1 2r

y_n x_n ð2rÞy_n for nK:

If we apply the Comparison Test 3.7.7 twice, we obtain the assertion in (a).

(b) Ifr¼0, then there existsK2N such that

0<x_ny_n for nK;

so that Theorem 3.7.7(a) applies. Q.E.D.

Remark The Comparison Tests 3.7.7 and 3.7.8 depend on having a stock of series that one knows to be convergent (or divergent). The reader will find that thep-series is often useful for this purpose.

3.7.9 Examples (a) The series X¹

n¼1

1

n²þn converges.

It is clear that the inequality 0< 1

n²þn< 1

n² for n2N is valid. Since the seriesP

1=n² is convergent (by Example 3.7.6(c)), we can apply the Comparison Test 3.7.7 to obtain the convergence of the given series.

(b) The series X¹

n¼1

1

n²nþ1 is convergent.

If the inequality

(10) 1

n²nþ1 1 n²

3.7 INTRODUCTION TO INFINITE SERIES 99

were true, we could argue as in (a). However, (10) isfalsefor alln2N. The reader can probably show that the inequality

0< 1

n²nþ1 2 n²

is valid for alln2N, and this inequality will work just as well. However, it might take some experimentation to think of such an inequality and then establish it.

Instead, if we takex_n:¼1=ðn²nþ1Þandy_n:¼1=n²;then we have xn

y_n ¼ n²

n²nþ1¼ 1

1ð1=nÞ þð1=n²Þ!1:

Therefore, the convergence of the given series follows from the Limit Comparison Test 3.7.8(a).

(c) The series X¹

n¼1

ffiffiffiffiffiffiffiffiffiffiffi1 nþ1

p is divergent.

This series closely resembles the series P 1= ffiffiffi

pn

, which is a p-series with p¼¹₂; by Example 3.7.6(e), it is divergent. If we letxn:¼1= ffiffiffiffiffiffiffiffiffiffiffi

nþ1

p andy_n:¼1= ffiffiffi pn

, then we have xn

y_n ¼ ffiffiffin p ffiffiffiffiffiffiffiffiffiffiffi nþ1

p ¼ 1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1þ1=n

p !1:

Therefore the Limit Comparison Test 3.7.8(a) applies.

(d) The series X¹

n¼1

1

n!is convergent.

It would be possible to establish this convergence by showing (by Induction) that n²<n!forn4, whence it follows that

0< 1 n!< 1

n² for n4:

Alternatively, if we let x:¼1=n!andy_n:¼1=n², then (whenn4) we have 0xn

y_n ¼n²

n!¼ n

12 ðn1Þ< 1 n2!0:

Therefore the Limit Comparison Test 3.7.8(b) applies. (Note that this test was a bit troublesome to apply since we do not presently know the convergence of any series for which the limit ofxn=y_nis really easy to determine.) _&

Exercises for Section 3.7

1. LetP

anbe a given series and letP

bnbe the series in which the terms are the same and in the same order as inP

anexcept that the terms for whichan¼0 have been omitted. Show that Panconverges toAif and only ifP

bnconverges toA.

2. Show that the convergence of a series is not affected by changing afinitenumber of its terms. (Of course, the value of the sum may be changed.)

3. By using partial fractions, show that (a) X¹

n¼0

1 nþ1

ð Þðnþ2Þ¼1; (b)X¹

n¼0

1 aþn

ð Þðaþnþ1Þ¼1

a>0;ifa>0: (c) X¹

n¼1

1

n nð þ1Þðnþ2Þ¼1 4: 4. IfP

xnandP

y_nare convergent, show thatP xnþy_n

ð Þis convergent.

5. Can you give an example of a convergent seriesP

xnand a divergent seriesP

y_nsuch that Pðxnþy_nÞis convergent? Explain.

6. (a) Calculate the value of P¹

n¼2

2=7

ð Þⁿ. (Note the series starts atn¼2.) (b) Calculate the value of P¹

n¼1ð1=3Þ²ⁿ. (Note the series starts atn¼1.) 7. Find a formula for the series P¹

n¼1r²ⁿwhenj jr <1.

8. Let r1;r2;. . .;rn;. . . be an enumeration of the rational numbers in the interval [0,1].

(See Section 1.3.) For a givene>0, put an interval of lengtheⁿabout thenth rational number r_nforn¼1;2;3;. . .;and find the total sum of the lengths of all the intervals. Evaluate this number fore¼0:1 ande¼0:01.

9. (a) Show that the series P¹

n¼1cosnis divergent.

(b) Show that the series P¹

n¼1

cosn

ð Þ=n²is convergent.

10. Use an argument similar to that in Example 3.7.6(f) to show that the series P¹

n¼1 1 ð Þ^pffiffi_nⁿ is convergent.

11. IfP

anwithan>0 is convergent, then isP

a²_nalways convergent? Either prove it or give a counterexample.

12. IfP

anwithan>0 is convergent, then isP ffiffiffiffiffipan

always convergent? Either prove it or give a counterexample.

13. IfP

anwithan>0 is convergent, then isP ffiffiffiffiffiffiffiffiffiffiffiffiffipananþ1

always convergent? Either prove it or give a counterexample.

14. If P

anwithan>0 is convergent, and ifbn:¼ða1þ þanÞ=nforn2N;then show that Pbnis always divergent.

15. Let P¹

n¼1

a nð Þbe such thatða nð ÞÞis a decreasing sequence of strictly positive numbers. Ifs nð Þ denotes thenth partial sum, show (by grouping the terms insð Þ2ⁿ in two different ways) that

1

2ðað Þ þ1 2að Þ þ þ2 2ⁿað Þ2ⁿ Þ sð Þ 2ⁿ að Þ þ1 2að Þ þ þ2 2ⁿ¹a2ⁿ¹

það Þ2ⁿ . Use these inequalities to show thatP¹

n¼1

a nð Þconverges if and only ifP¹

n¼1

2ⁿað Þ2ⁿ converges. This result is often called theCauchy Condensation Test; it is very powerful.

16. Use the Cauchy Condensation Test to discuss thep-series P¹

n¼1

1=n^p

ð Þforp>0.

17. Use the Cauchy Condensation Test to establish the divergence of the series:

(a) X 1

nlnn, (b) X 1

nðlnnÞðln lnnÞ,

(c) X 1

nðlnnÞðln lnnÞðln ln lnnÞ.

18. Show that ifc>1, then the following series are convergent:

(a) X 1

nðlnnÞ^c, (b) X 1

nðlnnÞðln lnnÞ^c.

3.7 INTRODUCTION TO INFINITE SERIES 101