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Let $a_n$ be a series. For any sub sequence $a_{n_k}$, exists sub sequence $a_{n_{k_l}}$ that converge to L.

Prove or disprove that $a_n\to L$.

My try: Let $a_{n_k}=(-1)^n$, thus $a_{n_k}$ has two sub sequences that converge (one to 1, one to -1), but $a_{n_k}$ has two partial sums, thus $a_n$ has at least two partial sums, therefore not converging to L.

I saw a proof of the statement, but it's not clear and there are some assumptions which are seem wrong to me.

Please prove or disprove the statement.

Thank you!

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marked as duplicate by Guy Fsone, Aqua, Giuseppe Negro, kccu, abiessu Dec 5 '17 at 17:44

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Assume that $(a_{n})_n$ does not converge to $L$. Then $L$ must have some neighbourhood $U$ such that $\forall n\exists k\geq n\; a_{k}\notin U$. Then a subsequence $(a_{n_{k}})_k$ with $a_{n_{k}}\notin U$ for each $k$ can be constructed. For every subsequence $(a_{n_{k_i}})_i$ of this sequence we have $a_{n_{k_i}}\notin U$ for each $i$, showing that it does not converge to $L$. This proves the statement.

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  • $\begingroup$ Great, thanks! may you explain why my try failed? $\endgroup$ – Galc127 Jan 22 '14 at 9:42
  • $\begingroup$ I think you are mixing up sequences $(a_n)_n$ and sums of sequences. There is no need to look at 'partial sums'. You start by saying that $a_n$ is a 'series' and then you speak of 'subsequences' of it. That makes it look inconsistent and difficult to judge. Also according to your definition of $a_{n_k}$ it does not depend on $k$ (but on $n$). It all needs some fixing. $\endgroup$ – drhab Jan 22 '14 at 9:48

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