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Given the sequence $\{a_n\}_{n=0}^\infty$ and the subsequences: $a_{3n}, a_{2n+1}, a_{2n}$ which converge. Prove that $a_n$ is a convergent sequence.

Thanks you very much.

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What have you tried? –  Alex Becker Apr 30 '12 at 17:43
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all I could think about was to use the fact that if a sequence is convergent then all of its subsequences that converge must covnerge to the same limit of the sequence itself; maybe I could assume that these subsequences converge to different limits and get a contradiction? –  Anonymous Apr 30 '12 at 17:50
    
Yes, you could approach it by contradiction, but there's an easier way. Let $L_{\text{odd}}=\lim_n a_{2n+1}$ and $L_{\text{even}}=\lim_n a_{2n}$, and use the convergence of $\langle a_{3n}\rangle$ to show that $L_{\text{odd}}=L_{\text{even}}$. –  Brian M. Scott Apr 30 '12 at 17:53

2 Answers 2

up vote 3 down vote accepted

Hint: first show that the three subsequences have the same limit. (The subsequence $(a_{3n})$ has a further subsequence that is a subsequence of $(a_{2n})$, for instance.)

Then note that given $n>1$, $a_n$ is a term of one of the subsequences $(a_{2n})$, $(a_{2n+1})$. (So, given $\epsilon>0$, choose $N$ so that for any $n\ge N$, each of $a_{2n}$ and $a_{2n+1}$ is within $\epsilon$ of the common limit. Then... .)

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"The subsequence $(a_{3n})$ has a further subsequence that is a subsequence of $(a_{2n})$" - that's clear; however, why can I deduce that therefore $(a_{3n})$ and $(a_{2n})$ have the same limit? –  Anonymous Apr 30 '12 at 18:06
    
@Anonymous Suppose $(a_{2n})$ converges to $L$. Then each of its subsequences converges to $L$. So some subsequence of $(a_{3n})$ converges to $L$. So $(a_{3n})$ converges to $L$ (since $(a_{3n})$ is convergent). –  David Mitra Apr 30 '12 at 18:09
    
OK, awesome; therefore all the three subsequences have the same limit. let $n>1$, I understand that $a_n$ is a term of the subsequences $(a_{2n})$, $(a_{2n+1})$. how do I continue from here? I know that if n is even I can write: $|a_{2n}-L|<\epsilon$ and if n is odd then $|a_{2n+1}-L|<\epsilon$; how can I deduce now that $|a_n-L|<\epsilon$? –  Anonymous Apr 30 '12 at 18:17
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@Anonymous Finishing up the parenthetical comment at the end of my answer: For $n\ge N$, both $a_{2n}$ and $a_{2n+1}$ are within $\epsilon$ of $L$. But then every term $a_{2N}$, $a_{2N+1}$, $a_{2N+2}$, $a_{2N+3}$, $\ldots\,$ is within $\epsilon$ of $L$. Note those latter terms are a tail of $(a_n)$; so, for the sequence $(a_n)$, if $n\ge 2N$ ... . –  David Mitra Apr 30 '12 at 18:32
    
finally got it; as always clear as crystal; thanks again for your great help. –  Anonymous Apr 30 '12 at 18:49

Note that it suffices to show that, $$ \lim a_{2n} = \lim a_{2n + 1}.$$ As $ a_{6n} $ is a subsequence of $a_{2n}$ and $a_{3n}$. We have $ \lim a_{2n} = \lim a_{3n}$ And as $a_{3n} = a_{2n_m + 1} $. We have $$\lim a_{2n + 1} = \lim a_{3n} = \lim a_{2n}. $$

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