Construction of conditionally convergent series with certain property Question: It is given that the series $\sum_{n=1}^\infty a_n$ is convergent, but not absolutely convergent and $\sum_{n=1}^\infty a_n=0$. Denote by $s_k$ the partial sum $\sum_{n=1}^k a_n$, $k=1,2,... $ 
then " it is possible"  that $s_k>0$ for all but a finite number of values of k ?  
My attempt: 
Define $$
s_k=\left\{\begin{array}{}
0&\text{if $k=2,3$ }\\[4pt]
\frac1{k}&\text{otherwise}
\end{array}\right.
$$
Then clearly $s_k>0$ for all but a finite number of values of k (as required)
Further, $s_k$ converges to $0$ and  hence $\sum_{n=1}^\infty a_n=0$. 
But $\sum_{n=1}^\infty a_n$ is not absolutely convergent (how can i prove it?)
Am i correct? Please help me

No, if $s_n$ decreases to $0$ after some finite point, let's call it $N,$ then we have
$$\sum_{n=N}^{\infty}(s_n-s_{n+1}) = s_N.$$
But the $n$th summand here is $-a_{n+1}$ and is nonnegative. It follows that  $\sum_{n=N}^{\infty}|a_{n+1}| < \infty,$ which is a contradiction.
I think the following idea works: We consider a series of the form
$$1-\frac{1}{2}+a_3 +\frac{1}{3}- \frac{1}{4}+a_6 +\frac{1}{5}- \frac{1}{6}+a_9+\cdots $$
Choose $a_3,a_6,\dots$ as follows
$$a_3 = \frac{1}{2}-1+\frac{1}{3!},$$ $$a_6 = \frac{1}{4}-\frac{1}{3}-\frac{1}{3!}+ \frac{1}{6!},$$ $$ a_9 = \frac{1}{6}-\frac{1}{5}-\frac{1}{6!} +\frac{1}{9!}, \dots$$
The idea is that $a_{3n}$ cancels out all previous summands, and then adds in the small amount of $1/(3n)!$ (which could be replaced by any positive sequence $\to 0.$) Note that $s_{3n}=1/(3n!)$ for all $n.$
.
