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This is a problem from ISI MStat 2018 PSA Problem 11 based on Sequence and subsequence.

Let \( {a_{n}}_{n \geq 1}\) be a sequence such that \( a_{1} \leq a_{2} \leq \cdots \leq a_{n} \leq \cdots\)

Suppose the subsequence \( {a_{2 n}}_{n \geq 1}\) is bounded. Then

- (A) \( \{a_{2 n}\}_{n \geq 1}\) is always convergent but \( \{a_{2 n+1}\}_{n \geq 1} \) need not be convergent.
- (B) both \( \{a_{2 n}\}_{n \geq 1}\) and \( \{a_{2 n+1}\}_{n \geq 1}\) are always convergent and have the same limit.
- (C) \( \{a_{3 n}\}_{n \geq 1}\) is not necessarily convergent.
- (D) both \( \{a_{2 n}\}_{n \geq 1}\) and \( \{a_{2 n+1}\}_{n \geq 1}\) are always convergent but may have different limits.

Sequence

Subsequence

But try the problem first...

Answer: is (B)

Source

Suggested Reading

ISI MStat 2018 PSA Problem 11

Introduction to Real Analysis by Bertle Sherbert

First hint

Given that \( a_{2n} \) is bounded . Again we have \( a_{1} \leq a_{2} \leq \cdots \leq a_{2n} \leq a_{2n+1} \leq a_{2n+2} \leq \cdots\) which shows that if \( a_{2n} \) is bounded then \( a_{2n+1} \) is also bounded .

Second Hint

Again both \( a_{2n} \) and \( a_{2n+1} \) both are monotonic sequence . Hence both converges .

Now we have to see whether they converges to same limit or not ?

Final Step

As both \( a_{2n} \) and \( a_{2n+1} \) are bounded hence \( a_{n} \) is bounded and it's already given that it is monotonic . Hence \( a_{n} \) converges . So, it's subsequences must converges to same limit . Hence option (B) is correct .

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