8.2Monotone Sequences

Definition 8.2.1 Let {tn}∞n=m be a given sequence. Then {tn}∞n=m is said to be

(a)increasing if tn < tn+1 for all n ≥ m;

(b)decreasing if tn+1 < tn for all n ≥ m;

(c)nondecreasing if tn ≤ tn+1 for all n ≥ m;

(d)nonincreasing if tn+1 ≤ tn for all n ≥ m;

(e)bounded if a ≤ tn ≤ b for some constants a and b and all n ≥ m;

(f)monotone if {tn}∞n=m is increasing, decreasing, nondecreasing or nonincreasing.

(g)a Cauchy sequence if for each > 0 there exists some M such that |an1 − an2 | < whenever n1 ≥ M and n2 ≥ M.

Theorem 8.2.1 (a) A monotone sequence converges to some real number if and only if it is a bounded sequence.

(b) A sequence is convergent if and only if it is a Cauchy sequence.

Proof.

Part (a) Suppose that an ≤ an+1 ≤ B for all n ≥ M and some B. Let L be the least upper bound of the sequence {an}∞n=m. Let > 0 be given. Then there exists some natural number N such that

L − < aN ≤ L.

Then for each n ≥ N, we have

L − < aN ≤ an ≤ L.

By definition {an}∞n=m converges to L.

Similarly, suppose that B ≤ an+1 ≤ an for all n ≥ M. Let L be the greatest lower bound of {an}∞n=m. Then {an}∞n=m converges to L. It follows that a bounded monotone sequence converges. Conversely, suppose that a

monotone sequence {an}∞n=m converges to L. Let = 1. Then there exists some natural number N such that if n ≥ N, then

|an − L| < − < an − L < L − < an < L + .

The set {an : m ≤ n ≤ N} is bounded and the set {an : n ≥ N} is bounded. It follows that {an}∞n=m is bounded. This completes the proof of Part (a) of the theorem.

Part (b) First, let us suppose that {an}∞n=m converges to L. Let > 0 be given. Then 2 > 0 and hence there exists some natural number N such that for all natural numbers p ≥ N and q ≥ N, we have

|ap − aq| = |(ap − L) + (L + aq)|

≤|ap − L| + |a1 − L|

<2 + 2

= .

It follows that {an}∞n=m is a Cauchy sequence.

Next, we suppose that {an}∞n=m is a Cauchy sequence. Let S = {an : m ≤ n < ∞}. Suppose > 0. Then there exists some natural number N such

It follows that S is a bounded set. If S is an infinite set, then S has some limit point q and some subsequence {ank }∞k=1 of {an}∞n=m that converges to q. Since > 0, there exists some natural number M such that for all k ≥ M, we have

322 CHAPTER 8. INFINITE SERIES

Also, for all k ≥ N + M, we get nk ≥ k ≥ N + M and

|ak − q| = |ak − ank + ank − q| ≤ |ank − ak| + |ank − q|

= .

It follows that the sequence {an}∞n=m converges to q. If S is a finite set, then some ak is repeated infinite number of times and hence some subsequences of {an}∞n=m converges to ak. By the preceding argument {an}∞n=m also converges to ak. This completes the proof of this theorem.

Theorem 8.2.2 Let {f(n)}∞n=m be a sequence where f is a di erentiable function defined for all real numbers x ≥ m. Then the sequence {f(n)}∞n=m is

(a)increasing if f0(x) > 0 for all x > m;

(b)decreasing if f0(x) < 0 for all x > m;

(c)nondecreasing if f0(x) ≥ 0 for all x > m;

(d)nonincreasing if f0(x) ≤ 0 for all x > m.

Proof. Suppose that m ≤ a < b. Then by the Mean Value Theorem for derivatives, there exists some c such that a < c < b and

f(b) − f(a) = f0(c), b − a

f(b) = f(a) + f0(c)(b − a).

The theorem follows from the above equation by considering the value of f0(c). In particular, for all natural numbers n ≥ m,

f(n + 1) = f(n) + f0(c),

for some c such that n < c < n + 1.

Part (a). If f0(c) > 0, then f(n + 1) > f(n) for all n ≥ m. Part (b). If f0(c) < 0, then f(n + 1) < f(n) for all n ≥ m. Part (c). If f0(c) ≥ 0, then f(n + 1) ≥ f(n) for all n ≥ m. Part (d). If f0(c) ≤ 0, then f(n + 1) ≤ f(n) for all n ≤ m.

This completes the proof of this theorem.

8.3Infinite Series

Definition 8.3.1 Let {tn}∞n=1 be a given sequence. Let

n

X s1 = t1, s2 = t1 + t2, s3 = t1 + t2 + t3, · · · , sn = tk,

k=1

for all natural number n. If the sequence {sn}∞n=1 converges to a finite number

L, then we write

∞

X

L = t1 + t2 + t3 + · · · = tk.

k=1

n

X

We call tk an infinite series and write

k=1

∞n

XX

We say that L is the sum of the series and the series converges to L. If a series does not converge to a finite number, we say that it diverges. The sequence {sn}∞n=1 is called the sequence of the nth partial sums of the series.

Theorem 8.3.1 Suppose that a and r are real numbers and a 6= 0. Then the geometric series

∞

a + ar + ar2 + · · · = Xark = 1 −a r,

k=0

if |r| < 1. The geometric series diverges if |r| ≥ 1.

Proof. For each natural number n, let

sn = a + ar + · · · + arn−1.

On multiplying both sides by r, we get

rsn = ar + ar2 + · · · + arn−1 + arn

Exercises 8.1

1.Define the statement that the sequence {an}∞n=1 converges to L.

2.Suppose the sequence {an}∞n=1 converges to L and the sequences {bn}∞n=1 converges to M. Then prove that

(a){can}∞n=1 converges to cL, where c is constant.

(b){an + bn}∞n=1 converges to L + M.

(c){an − bn}∞n=1 converges to L − M.

(d){anbn}∞n=1 converges to LM.

∞

(e)an converges to L , if M 6= 0. bn n=1 M

3.Suppose that 0 < an ≤ an+1 < M for each natural number n. Then prove that

(a){an}∞n=1 converges.

(b){−an}∞n=1 converges.

(c)akn ∞n=1 converges for each natural number k.