It is given that the series $ \sum_{n=1}^{\infty} a_n$ is convergent 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=1,2,3, \cdots $. Then


*

*$ s_k=0$ for infinitely many $k$

*$s_k>0$ for infinitely many $k$

*it is possible that $ s_k>0$ for all $k$

*it is possible that $ s_k>0$ for all but finite number of values of $k$.
Answer: 
Consider the sequence $ \{a_n \}$ defined by $ a_{2n-1}=\frac{1}{n}$ and $a_{2n}=-\frac{1}{n}$, so that
$ \sum_{n=1}^{\infty} a_n=1-1+\frac{1}{2}-\frac{1}{2}+\cdots $
Thus,
$ s_{2n-1}=\frac{1}{n} \to 0 \ as \ n \to \infty$,
$s_{2n} =0$
Thus,
$ \sum_{n=1}^{\infty} a_n=0$.
Also the series is not absolutely convergent.
Thus $s_{2n-1}=\frac{1}{n}>0$ for infinitely many $n$
hence option $(3)$ is true.
What about the other options?
help me
 A: Also option $4$ is true according to your example with a bit change where you need to swap the values of $a_1$ and $a_2$ together. The other options are also true according to your example.
A: Your example doesn't show that (3) is true, since there are infinitely many values where $s_k=0\not\gt 0$. 
However, if you did have an example to show that (3) was true, you would know the answers to all the other questions. If you can have a series satisfying (3), the same series satisfies (4). Also, by swapping signs in all the terms in your series, you get a counterexample to (1) and (2), so they are not necessarily true.
(The issue here is that (3) and (4) are statements that something that can happen, whereas (1) and (2) are statements that something must happen, so examples can prove (3) and (4), and disprove (1) and (2).)
So we want to find an example for (3). As a hint for this, try starting with the harmonic series and showing that you can insert signs in such a way that the sums $s_k$ are always positive, but less than say $2/k$. 
A: As has been said, if you have a correct example to (3), this solves all the other problems as well, (1) and (2) in the negative and (4) in the positive. My hint is to modify your example such that $s_{2n}$ forms a positive sequence that as a sum is (absolutly) convergent.
The latter part makes sure (which you need to prove) that the modified $a'_n$ is not suddenly becming absolutely convergent.
ADDED: A possible solution would be shooting for $s_{2n-1}=\frac1n$ and $s_{2n}=\frac1{2^n}$. This means $a_{2n}=\frac1{2^n} - \frac1n$ and $a_{2n-1}=\frac1n - \frac1{2^{n-1}}$ for $n > 1$ and $a_1=1$.
To prove that this sequence $(a_n)$ is not absolutely convergent, note that it is the sum of two parts: $p_1=(1,-1,\frac12,-\frac12,\frac13,-\frac13,\ldots)$ and $p_2=(0, \frac12,-\frac12,\frac14,-\frac14,\frac18,-\frac18,\ldots)$. 
$p_1$ is the sequence you used, and correctly noted as not absolutely convergent. $p_2$ is absolutely convergent. If they their sum $(a_n)$ was absolutely convergent, it would mean that $p_1 = (a_n) - p_2$ was absolutely convergent (sums/differences of absolutely convergent series' are also absolutely convergent), which isn't true.
