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

Check the Answer

Answer: is (B)

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 .

Similar Problems and Solutions

ISI MStat 2018 PSA Problem 11 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

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

Check the Answer

Answer: is (B)

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 .

Similar Problems and Solutions

ISI MStat 2018 PSA Problem 11 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

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