It helps to reformulate the assumptions in terms of $S_k$. We are told that
- $S_k\to 0$
- $\sum |S_{k+1}-S_k| =\infty$
and nothing else. Of course, there is nothing here that implies $S_k$ being zero, or positive for infinitely many values of $k$. The examples $S_k=(2+(-1)^k)/k$ and $S_k=(-2+(-1)^k)/k$ take care of all four parts, confirming Did's answer in a comment: "(a) and (b) are false in general while (c) and (d) hold".
I am gonna assume you want sequences which satisfy the given conditions individually. Because some of them are contradictory to each other. So you can't find a single sequence satisfying all conditions. One more thing, the sequence you considered for option $3$, won't work out since it doesn't converge to $0$, which is one of the requirement.
Consider $ \{1,\,-1,\,1/2,\,-1/2,\, 1/3,\,-1/3...\}$ the sum converges to $0$, and doesn't converge absolutely, and the sum is $0$ for Infinitely $k$, for even numbered groups.
Say if you were to swap every other pair, then the sequence would be $ \{1,\,-1,\,1/2,\,-1/2,\, -1/3,\,1/3, \,-1/4, \,1/4...\}$, then for some infinite $k$, sum is negative for other infinite $k$ it is positive. So option $1,\, 2$ are possible.
Take $ \{-log\, 2, \, 1\, , -1/2\, , 1/3\, ,-1/4\}$, sum will be positive for all $k$, except for first one. The sum still goes to zero, not absolutely convergent. So option $4$ is possible. It should easy to find out why option 3 is not possible taking the same example, and modifying it little bit.
Best Answer
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.