Uniform Convergence of Dirichlet Series $\sum_{n=1}^{\infty} (-1)^n/(n^z)$

Question

Show that the series $\sum_{n=1}^{\infty} (-1)^n/(n^z)$ converges uniformly on a region $\{z \in \mathbb{C} | Re(z) \geq \epsilon\}$ for $\epsilon>0$ and defines an analytic function on $\{z \in \mathbb{C} | Re(z)>0\}$..

Surely the $(-1)^n$ factor ensures the uniform convergence since the summands' modulus goes to zero but to actually show this has become tricky for me. I've thought about partial sums but it becomes real messy and the Dirichlet convergence theorem doesn't necessarily give uniform convergence

I'm not looking for a solution, instead a nudge in the right direction would be ideal.

Thanks in advanced for any help :)

Answer 1

Hint: $\sum\limits_{k=1}^{N}\frac {(-1)^{k}} {k^{z}} =\sum\limits_{k=1}^{N} s_k (\frac 1 {k^{z}}-\frac 1 {(k+1)^{z}}) +\frac {s_N} {(N+1)^{z}}$ (summation by parts) where $s_n=\sum\limits_{k=1}^{n} (-1)^{k}$. Use the boundedness of $(s_n)$ to show that $\sum\limits_{k=1}^{\infty} s_k (\frac 1 {k^{z}}-\frac 1 {(k+1)^{z}}) $ converges uniformly and $\frac {s_N} {(N+1)^{z}} \to 0$ uniformly on compact subsets of $\{z \in \mathbb C: Re(z) >0\}$.

Answer 2

It doesn't converge uniformly.

$$\eta(s) = \lim_{N \to \infty} \eta_N(s),\qquad \eta_N(s)=\sum_{n=1}^N (-1)^{n+1} n^{-s}, \Re(s) > 0$$ If the convergence was uniform on $\Re(s) \ge 1/2$, since $n^{-1/2-it}$ is bounded we would have that $\eta(1/2+it),t \in \Bbb{R}$ is bounded and $$\lim\sup_{T \to\infty} \frac1{2T}\int_{-T}^T |\eta(1/2+it)|^2 dt <\infty$$ But again by uniform convergence it would be $$ = \lim_{N \to \infty}\lim\sup_{T \to\infty} \frac1{2T}\int_{-T}^T |\eta_N(1/2+it)|^2 dt $$

$$= \lim_{N \to \infty} \lim\sup_{T \to\infty}\sum_{n=1}^N\sum_{m=1}^N (-1)^{m+1} (-1)^{n+1} \frac1{2T}\int_{-T}^T n^{-1/2-it}m^{-1/2+it} dt $$

$$= \lim_{N \to \infty} \sum_{n=1}^N\sum_{m=1}^N (-1)^{m+n} (nm)^{-1/2}\lim\sup_{T \to\infty}\frac1{2T}\int_{-T}^T (n/m)^{-it} dt $$

$$= \lim_{N \to \infty} \sum_{n=1}^N\sum_{m=1}^N (-1)^{m+n} (nm)^{-1/2} 1_{n/m=1}$$ $$ = \lim_{N \to \infty}\sum_{n=1}^N (-1)^{n+n} (nn)^{-1/2}= \infty$$

A contradiction.

Thus it doesn't converge uniformly.

A stronger result is that $\eta(1+\epsilon+it)$ is unbounded, which can be shown from the Euler product valid for $\Re(s) > 1$ $$\eta(s) = (1-2^{1-s})\exp(-\sum_{p \ prime} \log(1-p^{-s}))$$ and the fact the $\log p$ are $\Bbb{Q}$-linearly independent.