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
Sequence
Subsequence
But try the problem first...
Answer: is (B)
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 .
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
Sequence
Subsequence
But try the problem first...
Answer: is (B)
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 .
