# Sequence convergence and parentheses insertion

find an example for a series $a_{n}$ that satisfies the following:

1. $a_{n}\xrightarrow[n\to\infty]{}0$

2. ${\displaystyle \sum_{n=1}^{\infty}a_{n}}$ does not converges

3. There is a way to insert parentheses so ${\displaystyle \sum_{n=1}^{\infty}a_{n}}$ will converges.

I was thinking about the series:$1-1+\frac{1}{2}+\frac{1}{2}-\frac{1}{2}-\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}+...$

But I don't know how to prove 2.

Also will be nice to hear another examples, if any.

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Any sequence $a_n$ converging to zero such that there exists parenthesis with every term inside the parenthesis will work. You could replace your $2^n$ in the denominator by a $\log(n)$ or a $n^n$, it doesn't matter. As long as you use your trick and put enough brackets. =) – Patrick Da Silva Jun 2 '12 at 16:41
Look at the sequence of partial sums, $(S_n)$, defined by $S_n=\sum_{k=1}^n a_k$. It should be clear how to show that this sequence does not converge (find a subsequence that alternates between $0$ and $1$, e.g.). Recall that an infinite sum converges iff its sequence of partial sums converges. – David Mitra Jun 2 '12 at 16:56
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