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Let $(a_n)_{n=1}^\infty$ be a sequence in a metric space such that sub sequences $(a_{2n})_{n=1}^\infty$ , $(a_{2n+1})_{n=1}^\infty$, $(a_{3n})_{n=1}^\infty$ converge. Prove that the sequence $(a_n)_{n=1}^\infty$ converges.

I know there are similar questions but in those, subsequences converge to the same limit. Also I do not understand how $(a_{3n})_{n=1}^\infty$ will provide additional information about converge of $(a_n)_{n=1}^\infty$ since $(a_{2n})_{n=1}^\infty$ and $(a_{2n+1})_{n=1}^\infty$ cover $(a_{3n})_{n=1}^\infty$ . Is this question incomplete to do a proof or my way of thinking is wrong?

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    $\begingroup$ $(a_{3n})$ converging insures the other two have the same limit. $\endgroup$ – David Mitra Oct 30 '17 at 12:59
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    $\begingroup$ You need to prove that (i) if $\lim_{n \to \infty} a_{2n} = \lim_{n \to \infty} a_{2n + 1}$, then this is the limit of $(a_n)$, and (ii) use the convergence of $(a_{3n})$ to prove that the even and odd limits are equal. $\endgroup$ – AJY Oct 30 '17 at 12:59
  • $\begingroup$ Quite a neat problem that can be usually seen in every textbook :) It's quite intuitive if you think about it. The terms of $(a_{3n})$ alternate between terms of even and odd subsequences, so if they had different limit $(a_{3n})$ cannot be convergent. Also, if odd and even subsequences do converge to same limit, after sufficiently large $N$ all the terms of sequence will be very close to the limit $\endgroup$ – user340297 Oct 30 '17 at 13:15
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Hint. Note that $(a_{6n})_{n=1}^\infty$ is a subsequence of $(a_{2n})_{n=1}^\infty$ and $(a_{3n})_{n=1}^\infty$. Moreover $(a_{6n+3})_{n=1}^\infty$ is a subsequence of $(a_{2n+1})_{n=1}^\infty$, and $(a_{3n})_{n=1}^\infty$. Recall that any subsequence of a converging sequences is convergent to the same limit.

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