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?
$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{\alpha}} = \sum_{m\ge 1} \int_{2m-1}^{2m} \alpha x^{-\alpha-1}dx$$ $$ = \frac12+\frac{\alpha}2 \sum_{m\ge 1} (\int_{2m-1}^{2m} x^{-\alpha-1}dx-\int_{2m}^{2m+1} x^{-\alpha-1}dx) $$ It suffices to prove that the latter series is bounded on $[0,1]$.