Prove that $ \lim_{\alpha\to 0^+}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{\alpha}}=1/2 $. Recently I came cross an interesting problem as follows.

Prove that $ \lim_{\alpha\to 0^+}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{\alpha}}=1/2 $.

This makes me think of the analytic extension of zeta function. But I do know how to prove it. Can you give me some hints?
 A: $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{\alpha}} = \sum_{m\ge 1} \int_{2m-1}^{2m} \alpha x^{-\alpha-1}dx$$ $$ =
\frac12+\frac{\alpha}2 \sum_{m\ge 1} (\int_{2m-1}^{2m} x^{-\alpha-1}dx-\int_{2m}^{2m+1} x^{-\alpha-1}dx)
 $$
It suffices to prove that the latter series is bounded on $[0,1]$.
A: A good method to evaluate the sum (and prove the limit) is to group successive terms together in such a way that convergence is guaranteed.
$$S = \frac {1}{1^\alpha}-\frac {1}{2^\alpha}+\frac {1}{3^\alpha}-\frac {1}{4^\alpha}+\frac {1}{5^\alpha}- ...$$
Rewrite this as follows:
$$S = (1/2)\frac {1}{1^\alpha} + (1/2)[\frac {1}{1^\alpha}-\frac {2}{2^\alpha}+\frac {1}{3^\alpha}]+(1/2)[\frac {1}{3^\alpha}-\frac {2}{4^\alpha}+\frac {1}{5^\alpha}]+...$$
Now it is straightforward to demonstrate that the terms $[\frac {1}{(n-1)^\alpha}-\frac {2}{n^\alpha}+\frac {1}{(n+1)^\alpha}]$ are of order $n^{-2}$ for large $n$. Hence their sum converges to a finite value. Secondly, these terms are linear in the parameter $\alpha$. So taking the limit of $\alpha$ to zero is straightforward. The first term in $S$ trivially yields $1/2$, whereas the higher terms do not contribute.
