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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

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

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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

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

Subscribe to Cheenta at Youtube


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