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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 be a sequence such that Suppose the subsequence is bounded. Then

• (A) is always convergent but need not be convergent.
• (B) both and are always convergent and have the same limit.
• (C) is not necessarily convergent.
• (D) both and 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 is bounded . Again we have which shows that if is bounded then is also bounded .

Again both and both are monotonic sequence . Hence both converges .

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

As both and are bounded hence is bounded and it's already given that it is monotonic . Hence 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 be a sequence such that Suppose the subsequence is bounded. Then

• (A) is always convergent but need not be convergent.
• (B) both and are always convergent and have the same limit.
• (C) is not necessarily convergent.
• (D) both and 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 is bounded . Again we have which shows that if is bounded then is also bounded .

Again both and both are monotonic sequence . Hence both converges .

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

As both and are bounded hence is bounded and it's already given that it is monotonic . Hence converges . So, it's subsequences must converges to same limit . Hence option (B) is correct .

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