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 May 14th, 2017, 11:51 PM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 106 Thanks: 1 Convergence of a sub sequence Hello Any one help me with this!!! If ${a_n }$ is a sequence converging to $l$. Let $b_n$ = $a_2n$, if n is odd, $a_3n$, if n is even. Then the sequence ${b_n}$ A. need not converge B. should converge to $0$. C. should converge to $2l$ or to $3l$. D. should converge to $l$. I know that if a sequence is convergent then its subsequents also converge. But I dont know whether they converge to the same limit as that of the original sequence or not ? Please help!!! thanks in advance
 May 15th, 2017, 03:24 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,405 Thanks: 611 Yes, of course they have to converge to the same limit! If the sequence, $a_n$, converges to x, that means, for any $\epsilon> 0$ and sufficiently large n, $|a_n- x|< \epsilon$. Now, suppose some subsequence, $a_{f(n)}$ converged to some other number, y. That would mean that, for some $\epsilon> 0$ and sufficiently large n, $|a_{f(n)}- y|< \epsilon$. Take $\epsilon< |x- y|/2$ and realize that those cannot both be true!
May 15th, 2017, 03:39 AM   #3
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Quote:
 Originally Posted by Country Boy Yes, of course they have to converge to the same limit! If the sequence, $a_n$, converges to x, that means, for any $\epsilon> 0$ and sufficiently large n, $|a_n- x|< \epsilon$. Now, suppose some subsequence, $a_{f(n)}$ converged to some other number, y. That would mean that, for some $\epsilon> 0$ and sufficiently large n, $|a_{f(n)}- y|< \epsilon$. Take $\epsilon< |x- y|/2$ and realize that those cannot both be true!
Thank you so much for the clarification

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