Different limits for the alternating harmonic series? Show that the series 
$$\sum_{n=1}^{\infty} \dfrac{(-1)^n}{n}$$
is not absolutely convergent. Show that by permuting the terms of the series one can obtain series with different limits. 
I am able to show that this is not absolutely convergent. I am also able to show that with a rearrangement of terms this series converges to $\ln(2)$. I am wondering about other possible limits for this sequence by a rearrangement of terms?
Also this is not a homework problem. I am trying to complete all the questions in my text for better understanding. Any help and comments are appreciated. Thank you.  
 A: A theorem of Riemann says you can permute the terms and get any limit if you want, if the series is convergent and not absolutely convergent. I encourage you to try to prove it. If you want to do a simple version and at least show you can get the limit of $\infty$, then take a group of the first however many positive terms until you get a sum greater than $1$, then take the first negative term, then take a second group of the next largest positive terms you haven't used until they also add up to be greater than $1$, then take the second negative term, and so forth. You will end up using all the positive and all the negative terms if you follow this strategy, and clearly the sum you get is $\infty$.
A: As others have mentioned, there's a theorem that says in fact you can rearrange it to any limit - even plus or minus infinity.
However, if you just want a single different limit, try
$$
\frac12 + \frac14 - \frac11 + \frac16 + \frac18 - \frac13 + \frac1{10} + \frac1{12} - \frac15 + \cdots
$$
Which you should be able to show approaches $-\frac12 \ln 2$.
Note that the series you give ($-1 + \frac12 - \frac13 + \cdots$) actually converges to $- \ln 2$ not $\ln 2$, so perhaps you meant to say you can show it rearranges to $-\ln 2$.
A: One of the simplest ways I have rearranged series is the following: if $n=2k+1$ for some $k,$ switch $n$ with $4k+2,$ and vice versa. Leave all other $n$ fixed. The new sum looks like 
$$-\frac{1}{2}+1-\frac{1}{6}-\frac{1}{4}-\frac{1}{10}+\frac{1}{3}-\frac{1}{14}-\frac{1}{8}...$$
Try to show that this sum converges to $-\frac{3}{4}\log(2).$
One possible way is to show that the new sum has the same limit as the sum
$$\sum_{n=0}^\infty -\frac{1}{8n+2}+\frac{1}{2n+1}-\frac{1}{8n+6}-\frac{1}{4n+4},$$ if finding this limit is easier.
