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!


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.

  • $\begingroup$ 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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