Question: Let {an} be a sequence and L  is any real number. Prove that {an} =1 converges to L  if and only if every monotone subsequence of {an} converges to L .

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Question: Let {an} be a sequence and L  is any real number. Prove that {an} =1 converges to L  if and only if every monotone subsequence of {an} converges to L .

 

Tip:  prove the contrapositive by using a) in image, and the Bolzano-Weierstrass Theorem.

(a) Let {an} be any sequence and L any real number. Prove that if {a,} does
not converge to L, then there exists an e> 0 and a subsequence {a} such
that an - L > e for all k.
Tip: Start by writing down what it means that {am} does not converge to L,
i.e. write down the formal negation of the statement {a,} converges to L. Use
this formal negation and induction on k to construct an increasing sequence of
natural numbers {ng} such that |an - L| > e for all k.
3D1
Transcribed Image Text:(a) Let {an} be any sequence and L any real number. Prove that if {a,} does not converge to L, then there exists an e> 0 and a subsequence {a} such that an - L > e for all k. Tip: Start by writing down what it means that {am} does not converge to L, i.e. write down the formal negation of the statement {a,} converges to L. Use this formal negation and induction on k to construct an increasing sequence of natural numbers {ng} such that |an - L| > e for all k. 3D1
Expert Solution
Step 1

Consider a sequence, an and L be any real number.

To prove that, an converges to L if and only if every monotone sub-sequence of an converges to L.

Suppose that an converges to L.

Since every convergent sequence is bounded, an is bounded.

Then by Bolzano-Weierstrass Theorem,

A bounded sequence of real numbers has a convergent sub-sequence.

Let ank is a monotone sub-sequence of an

Since, an converges to L, by definition, 

For every ε>0, there exists a positive integer m such that, 

an-L<ε   nm

In particular, 

ank-L<ε   nm

Therefore, ank converges to L.

So, if an converges to L, then every monotone sub-sequence of an converges to L.

 

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