Sequence convergence and parentheses insertion find an example for a series $a_{n}$
  that satisfies the following:


*

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

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

*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.

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. 
