# convergent but not absolutely convergence series problem which has zero sum

It is given that the series $\sum_{n=1}^\infty a_n$ is convergent, but not absolutely and $\sum_{n=1}^\infty a_n=0$. Denote by $S_k$ the partial sum $\sum_{n=1}^k a_n$ , $k=1,2,\dots$ Then,
(a) $S_k=0$ for infinitely many $k$;
(b) $S_k>0$ for infinitely many $k$ , $S_k<0$ for infinitely many $k$;
(c) it is possible that $S_k>0$ for all $k$;
(d) it is possible that $S_k>0$ for all but finite number of values of $k$.

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Not sure I know where the answer below and its comment are aiming at, but anyway, (a) and (b) are false in general while (c) and (d) hold. –  Did Dec 16 '12 at 19:31

Hint: Take the sequences $$a_n=\begin{cases}\frac{1}{n}&\mbox{if, }n=2,4,6,...\\ \frac{-1}{n+1}&\mbox{if, }n=1,3,5,...\end{cases}$$ and $$b_n=\begin{cases}\frac{-1}{n}&\mbox{if, }n=2,4,6,...\\ \frac{1}{n+1}&\mbox{if, }n=1,3,5,...\end{cases}$$ The series $$\sum_{n=1}^{\infty}a_n=0=\sum_{n=1}^{\infty}b_n$$ do not converge absolutely. This should help you rule out some options and point you in the right direction.

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from your example it is clear that a,b are true.but how can i verify d. –  gumti Dec 16 '12 at 16:06
also i think c is false as it gives the summation of the series must greater than 0 –  gumti Dec 16 '12 at 16:07
@gumti yes c is false –  Nameless Dec 16 '12 at 16:08
why a false? for those examples the partial sum would be 0 if k is even –  gumti Dec 16 '12 at 16:10
@gumti Yeah sorry. I was thinking about $a_n$ not $S_n$. (d) is easily verfied if you choose a sequence slightly different. –  Nameless Dec 16 '12 at 16:11

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".

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