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# ISI MStat 2018 PSA Problem 11 | Sequence & it's subsequence

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

## Sequence & it's subsequence - ISI MStat Year 2018 PSA Problem 11

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.

### Key Concepts

Sequence

Subsequence

ISI MStat 2018 PSA Problem 11

Introduction to Real Analysis by Bertle Sherbert

## Try with Hints

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 .

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 ?

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

## Sequence & it's subsequence - ISI MStat Year 2018 PSA Problem 11

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.

### Key Concepts

Sequence

Subsequence

ISI MStat 2018 PSA Problem 11

Introduction to Real Analysis by Bertle Sherbert

## Try with Hints

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 .

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 ?

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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