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Let series $\sum^\infty a_n$ is convergent but not absolutely convergent. And $\sum^\infty a_n =0$. Denote $ s_k $ the partial sum $\sum_{n=1}^k a_n $, $k=1,2,\dots$ then which of following ARE true.

1.$ s_k=0$ for infinitely many $k$.

  1. $ s_k \gt 0$ for infinitely many $k$, and $ s_k \lt 0$ for infinitely many $k$

  2. It is possible that $ s_k \gt 0$ for all $k$

  3. It is possible that $ s_k \gt 0$ for all but finite number of values of $k$.

I think series could be alternating one, converging to zero. So its partial sum sequence converges to zero. The partial sum sequence won't be monotonic, because it's not a positive term series, that's why I think option 2 is true and option 3 should be false. but about other options I don't know. Whats should be the correct options?

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  • $\begingroup$ 3. is possible. Start with $a_1=1$. Take the next $9$ terms to be $-.1$. Take the $11$'th term to be $.4$, to give a partial sum of $1/2$. Then take small negative terms till you get close to $0$. Then take the next term approximately $1/3$, to give a partial sum of $1/3$. And so on. $\endgroup$ – David Mitra Jan 29 '15 at 12:52
  • $\begingroup$ Whst about 1 and 4, I think atleast they cant be true. Isn'Isn't it $\endgroup$ – singularity Jan 29 '15 at 13:03
  • $\begingroup$ In my last comment, rather than "taking a bunch of small steps back to $0$", you could just take one big step. 4. can hold since 3. can. 1. and 2. need not hold since 3. can hold. $\endgroup$ – David Mitra Jan 29 '15 at 13:47
  • $\begingroup$ Is there a reason why 2 cannot hold? $\endgroup$ – singularity Jan 29 '15 at 13:51
  • $\begingroup$ A series satisfying 3. would not satisfy 2. So 2. is false in general. There are series, though, that satisfy 2. $\endgroup$ – David Mitra Jan 29 '15 at 15:15
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Let us consider

$$s_k=\sum_{n=1}^k\frac {(-1)^{n+1}} n$$

This series is called the alternating harmonic series and it is a canonical example of a series that is convergent but not absolutely convergent.

So first $s_k$ is convergent (this is a well known result, you can try to group the terms two-by-two), but not absolutely convergent (harmonic series, also a classic result).

  1. $\forall k\in\mathbb N,s_k>0$ (intuitively, at each step, either you increase $s_k$ or you decrease by less than what you added at the previous step), so 1. is false in general
  2. Same reasoning the for 1., false in general
  3. True, we just gave an example
  4. Also true, you can take any series verifying 3., shift it by any finite number of steps and set the first terms to 0
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