Sum of a series in real analysis It is given that the series $\sum\limits_{n=1}^{\infty}a_n$ is convergent but not absolutely convergent and $\sum\limits_{n=1}^{\infty}a_n=0$. Let $s_k=\sum\limits_{n=1}^{k}a_n,k=1,2,...$. Then


*

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

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

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

*It is possible that $s_k>0$ for all but a finite number of values of $k$
In my opinion 1. and 2. are correct options but 3. is incorrect as the series is not absolutely convergent, therefore all the terms of the series are not positive. But what about 4? I strongly feel that it is incorrect. But how? Please help!
 A: It is possible for 3. to be satisfied.
Consider the series
$$S=
(1+1/2) -1 +(1/2-1/3) -1/2 +(1/2-1/9)-1/3+(1/4-1/27)-1/4+\cdots.
$$
(so $a_1= 1+1/2$, $a_2=-1$, $a_3=1/2-1/3$, $\ldots$).
Here 
$\ \ \ \ \ s_{2n} = {1\over 2} -\sum\limits_{k=1}^{n-1} (1/3)^k$
and
$\ \ \ \ \ s_{2n+1} = {1\over 2} -\sum\limits_{k=1}^{n-1} (1/3)^k  +{1\over n+1}-(1/3)^{n } =\Bigl( {1\over2} -\sum\limits_{k=1}^n(1/3)^k\Bigr)+{1\over n+1}.$
Both $(s_{2n}) $ and $(s_{2n+1})$ are sequences of positive terms with limit $0$.
So, $S$ converges to $0$ and all its partial sums are positive. $S$, however, is not absolutely convergent.
A: It is "clear" that $s_k$ need not ever be equal to 0. So if justification is not asked for, things are easy. However, for certainty we need to provide an example. 
It is a standard fact that the series 
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$$
is conditionally convergent and has sum $\ln 2$. It is not hard to show that $\ln 2$ is irrational. 
We modify the series so that it has sum $0$. Keep the positive terms. Replace $-\frac{1}{2}$ by $-\left(\frac{1}{2}+\frac{\ln 2}{2^1}\right)$, replace $-\frac{1}{4}$ by $-\left(\frac{1}{4}+\frac{\ln 2}{2^2}\right)$, replace $-\frac{1}{6}$ by $-\left(\frac{1}{6}+\frac{\ln 2}{2^3}\right)$, and so on. 
It is not difficult to show that the modified series is conditionally convergent and has sum $0$.
However, partial sums are never equal to $0$. For any partial sum other than the first is of the shape $r+s\ln 2$, where $r$ and $s$ are rational and $s\ne 0$. So any partial sum after the first is irrational, and in particular cannot be $0$. 
