# An example to divergent sequence that there exists parenthesization such that it converges upon

I'm asked to find an example of a diverging sequence $\sum_{n\in\mathbb{N}}a_n$ such that $\lim_{n\to\infty} a_n = 0$ but there exists parenthesization such that $\sum_{n\in\mathbb{N}}a'_n<\infty$ ($a'_n$ is the parenthesised sequence).

Clearly, we need to find a "forbidden" parenthesization that will change the convergence status of the sequence.

We learned that parenthesization is allowed in 2 cases:

1. If the general term of the sequence $\to_{n\to\infty}{0}$ and the number of terms in each parentheses is bounded.
2. All the terms in each parentheses are of the same sign.

Couldn't find one so far... Will appreciate your help.

Thanks.

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Let $s_n = \sum_{i=1}^n$ be the sequence of partial sums of your series. "Parenthesizing" amounts to picking a subsequence of $s_n$. However, if $s_n$ diverges, say $s_n \rightarrow \infty$, then no subsequence can be convergent.
Perhaps the original series isn't meant to be divergent but rather non-convergent? In this case, you can take $1,-1,1/2,1/2,-1/2,-1/2,1/3,1/3,1/3,-1/3,-1/3,-1/3,\ldots$.
Thanks. Why do we need $2n$ terms in each parentheses? –  Amihai Zivan May 29 '12 at 13:58
The reason is that you want $a_n\rightarrow 0$. Otherwise you could take $1,-1,1,-1,\ldots$. –  Yuval Filmus May 29 '12 at 21:57