Variation of the Riemann Zeta Function

Question

We know that the Riemann Zeta function $$\sum_{n=1}^{\infty} \frac{1}{n^s}$$ converges when $Re(s) > 1$. How do you prove the convergence of the series $$\sum_{n=1}^{\infty} \frac{(-1)^n}{(z+n)^s}$$ converges when $Re(s) > 0$ and $z\in \mathbb{C}$ except for some $z$. Any idea what restriction we need to have on $z$ besides $z$ cannot be negative integer?

Answer 1

You asked the same question there

If $z$ is a negative integer then $(z+n)^{-s}$ is undefined.

Thus we assume it is not. For $\Re(s) > 1$ it converges absolutely. For $\Re(s) > 0$ we need a partial summation

$$\sum_{n=1}^N (-1)^n (z+n)^{-s} = (\sum_{n=1}^N (-1)^n) (z+N)^{-s}+\sum_{n=1}^{N-1}(\sum_{m=1}^n (-1)^m) ((z+m)^{-s}-(z+m+1)^{-s})$$ The first term $\to 0$, and for the second term we use that $$(z+m)^{-s}-(z+m+1)^{-s} = \int_0^1 s(z+x)^{-s-1}dx= O(s(z+m)^{-s-1})$$ Finally we let $N\to \infty$.