Can a series be rearranged so that it's subsequences tend to two different values? Prove that there exists a rearrangement of $\sum\limits_{n=1}^\infty\frac{(-1)^n}{n}$ such that the partial sums of subsequences tend to both $1$ and $-1$. This is different than asking how the series can be rearranged such that it tends to $1$ or $-1$.
 A: Outline: The sum of the positive terms diverges, as does the sum of the negative terms.  So if we start anywhere, the sum of the remaining positive terms diverges, as does the sum of the remaining negative terms. 
Add together positive terms, in the normal order, until the sum gets above $1$. Then add negative terms until the sum dips below $-1$. Then add fresh positive terms until the sum gets above $1$. Continue. It is not hard to see that $1$ and $-1$ are accumulation points of the partial sums.   
A: As $\sum_{n \ge 1} \frac{1}{2 n}$ and $\sum_{n \ge 0} \frac{1}{2 n + 1}$ both diverge, you can play  the following game for any $\alpha \in \mathbb{R}$ you care to pick:


*

*Add enough terms from the first series to just go over $\alpha$

*Substract enough terms from the second to just fall below $\alpha$

*Add enough from the following terms of the first series to go over $\alpha$

*...


The resulting series contains all terms of the original, just shuffled. It clearly converges to $\alpha$.
A: Generally, here applies the Riemann series theorem, which states that :

If the series $\sum\limits_{n=1}^\infty a_n$ converges, but it does
  not converge absolutely, i.e. $\sum\limits_{n=1}^\infty
> |a_n|=+\infty$, then you can rearrange the terms series with suitable
  ways, so that you can obtain:
$$\sum\limits_{n=1}^\infty a_{\sigma(n)}=L$$for all $L\in \mathbb
> R\cup\{-\infty,+\infty\}$. (where $\sigma(n)$ is a permutation of the
  terms of the sequence)

This, applied to your case, imples that since $\sum\limits_{n=1}^\infty \left|\frac{(-1)^n}{n}\right|=\infty$, you can rearrange the terms of the sequence, so that the partial sums converge to any number you want.
Actually, there is no general method to find the desired permutation, so that the partial sums converge to -1, or 1. Thus, you should use your inspiration, and combine some terms in the begining with some terms later in the series, in order to produce some zero's.
