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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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    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 $L_{\text{odd}}=L_{\text{even}}$.2012-04-30

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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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    @Anonymous You're welcome. Glad to help.2012-04-30
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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}. $