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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

The series $$1-1+\frac{1}{2}+\frac{1}{2}-\frac{1}{2}-\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}+...$$ is indeed divergent, because it has infinitely many partial sums equal to $1$, and infinitely many partial sums equal to $0$. Inserting parentheses as $$\left(1-1\right)+\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}\right)+...$$ yields a convergent series.

More generally: take any divergent series in which $\limsup S_N\ge 0$ and $\liminf S_N\le 0$ ($S_N$ are partial sums). Split every term into several, to ensure $a_n\to 0$. Insert $k$th closing parenthesis when the partial sum drops below $1/k$ in absolute value.

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D'Alembert's ratio test will help you with nr. 2.

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