# partial sum of series which is convergent but not absolutely convergent

I am stuck on the following exercise.

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,... Then

1. $s_k=0$ for infinitely many k

2. $s_k>0$ for infinitely many k, and $s_k<0$ for infinitely many k

3. it is possible that $s_k>0$ for all k

4. it is possible that $s_k>0$ for all but a finite number of values of k

Here $\sum_{n=1}^\infty a_n=\lim_{k\to \infty}s_k=0$ hence it is possible that its partial sum $s_k=0$ for infinitely many k, hence first option is correct.

Here series is convergent but not absolutely convergent therefore this series has negative terms hence $s_k$ can be greater than and less than $0$ infinitely many times ,hence option 2 is correct.
I have example $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$ this series is convergent but not absolutely convergent and $s_k>0$ for all k for this series, hence option 3 is correct.
I am not getting option 4.

please correct me if i am wrong.

Thank you.

• Are you trying to prove that all four options happen all the time? that each of the four options can happen some of the time? that one of the options must happen all of the time? Please clarify. – Gerry Myerson Nov 4 '16 at 6:24
• i am trying to prove that each of the four options can happen some of the time. sorry for late response – dipali mali Nov 8 '16 at 10:16

## 2 Answers

1. Take $a_n=\dfrac{(-1)^n}{n}$.

Then $s_1=-1,s_2=\frac{-1}{2}$,...

We have $s_n<0$ for all $n$. So A is false .

1. B is false also

3.Take $a_n=\dfrac{(-1)^{n+1}}{n}$.

Then $s_n>0$ for all $n$.

• The examples are supposed to satisfy $\sum a_n=0$. – Gerry Myerson Nov 8 '16 at 11:59

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.

• You are saying option 3 is not possible? – Gerry Myerson Nov 8 '16 at 12:00
• Yes. I came up with counterexamples for other options – jnyan Nov 8 '16 at 12:20
• Why can't the partial sums be $1,1/10,1/2,1/100,1/3,1/1000,1/4,1/10000,\dots$? – Gerry Myerson Nov 8 '16 at 20:09
• For which option? – jnyan Nov 9 '16 at 3:27
• I am sorry. The third option is also possible – jnyan Nov 9 '16 at 16:10