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 :)

  • $\begingroup$ It converges uniformly for $\Re(z) \ge 1+\epsilon$, for $\Re(z) \ge \epsilon$ it only converges locally uniformly : $\sum_{n=1}^\infty (-1)^n n^{-z} = -z\sum_{n=1}^\infty \int_{2n-1}^{2n} x^{-z-1}dx$ $\endgroup$
    – reuns
    Jun 7, 2019 at 21:00
  • $\begingroup$ That the sequence is alternating per the numerator is not particularly relevant as the imaginary component to $z$ will cause $k^z$ to move around on the complex plane. $\endgroup$
    – user317176
    Jun 7, 2019 at 21:25
  • 1
    $\begingroup$ This is the negative of the Dirichlet eta function. $\endgroup$
    – Somos
    Jun 7, 2019 at 22:55

2 Answers 2


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 $\{z \in \mathbb C: Re(z) >0\}$.

  • $\begingroup$ Than you very much. Does this mean the first üart about uniform convergence for all z with real part greater than or equal to some epsilon greater than zero is wrong? $\endgroup$
    – MrHolmes
    Jun 8, 2019 at 0:07
  • $\begingroup$ The convergence is uniform for $Re(z) \geq \epsilon$ and $|z|$ bounded. But this is enough to say that the sum of the series is analytic in the right half plane. $\endgroup$ Jun 8, 2019 at 0:13
  • $\begingroup$ Why don't you say that the convergence is not uniform, only locally uniform.. ? Also without $\int_k^{k+1} z t^{-z-1}dt$ it is not obvious that $\sum\limits_{k=1}^{\infty} s_k (\frac 1 {k^{z}}-\frac 1 {(k+1)^{z}})$ converges for $z$ non-real $\endgroup$
    – reuns
    Jun 8, 2019 at 1:49
  • $\begingroup$ @reuns I have posted hints, not a complete answer. If OP had asked for more details I would have given. $\endgroup$ Jun 12, 2019 at 5:15
  • $\begingroup$ I didn't write my answer for the OP but for you : the title is the uniform convergence of $\eta(s)$, it doesn't converge uniformly and it is not really trivial to show it $\endgroup$
    – reuns
    Jun 12, 2019 at 15:18

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.


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