$\sum a_n$ be convergent but not absolutely convergent, $\sum_{n=1}^{\infty} a_n=0$ Let $\sum a_n$ be convergent but not absolutely convergent, $\sum_{n=1}^{\infty} a_n=0$,$s_k$ denotes the partial sum then could anyone tell me which of the following is/are correct?
$1.$ $s_k=0$ for infinitely many $k$
$2$. $s_k>0$ and $<0$ for infnitely many $k$
3.$s_k>0$ for all $k$
4.$s_k>0$ for all but finitely many $k$
if we take $a_n=(-1)^n{1\over n}$ then $\sum a_n$ is convergent but not absolutely convergent,but I don't know $\sum_{n=1}^{\infty} a_n=0$? so I am puzzled could any one tell me how to proceed?
 A: None of them are necessarily true.
We can easily compute a series from its partial sums, so let's specify the $s_k$.
Define
$$
s_k=\left\{\begin{array}{}
-\frac1k&\text{if $k$ is odd}\\[4pt]
-\frac1{k^2}&\text{if $k$ is even}
\end{array}\right.
$$
Then $a_1=-1$ and for $k\gt1$,
$$
a_k=\left\{\begin{array}{}
\frac1{(k-1)^2}-\frac1k&\text{if $k$ is odd}\\[4pt]
\frac1{k-1}-\frac1{k^2}&\text{if $k$ is even}
\end{array}\right.
$$
Show that this series is not absolutely convergent, its sum is $0$, and it fails to satisfy any of the conditions.
A: It is easy to see that the first statement is wrong. The idea for a counterexample is in 3.
For the second statement use Riemann's rearrengement theorem, which (if you unterstood the proof) gives you the existence of such a series.
For the third statement construct $a_k$ such that $s_k=(-1)^k \cdot \frac{1}{k}$.
The fourth look at the third.
On the other hand all of the things can be true.
For the first let $(b_n)_{n\in \mathbb{N}}$ be an arbitrary null sequence. Define $a_n$ via
\[a_n= \begin{cases} b_k & 2k=n\\
-b_k & 2k+1=n
\end{cases} \]
We see that 
\[  \sum_{n=0}^\infty a_n = 0\]
and furthermore 
\[\sum_{n=1}^{2N+1} a_n=0\]
for any $N\in \mathbb{N}$. If you chose a "slow" enough null sequence it won't be absolute convergent.
For the second to be true use the series constructed in the first part at 3.
For the third part make something that a subsequence of $s_k$ is $\frac{1}{k}$, and blow up the rest with $n$ terms with values $\pm \frac{1}{n}$ and let $n \to \infty$ while $k\to \infty$. 
The fourth is solved by the third.
