Differentiability of the Riemann function

Question

I have defined the function $\zeta : (1, +\infty) \longrightarrow \mathbb{R}$ defined as: $$\zeta(x) = \sum_{n=1}^{+\infty} \frac{1}{n^x}$$

I have to study the differentiability of this function. I have tried to use the theorem that allows you to swap summation and derivative, but I don't know how to prove that $\sum_{n=1}^{+\infty} g'(x)$ converges uniformly, necessary to use the theorem, where $g(x) := \frac{1}{n^x}$

Answer 1

Let $\zeta(x)=\sum_{n=1}^\infty \frac{1}{n^x}$ be a series representation for the Riemann-Zeta Function for $x\in (1,\infty)$.

Let $z'(x)$ be the series of the differentiated terms of the Riemann Zeta function so that

$$z'(x)=-\sum_{n=1}^\infty \frac{\log(n)}{n^x}\tag 1$$

for $x\in(1,\infty)$. For all $\delta>1$, the series in $(1)$ converges uniformly for $x\ge \delta$.

Therefore, for all $x\ge \delta>1$, we have

$$\begin{align} \zeta'(x)&=\frac{d}{dx}\sum_{n=1}\frac{1}{n^x}\\\\ &=-\sum_{n=1}^\infty \frac{\log(n)}{n^x}\tag2 \end{align}$$

Since the equality in $(2)$ is true for all $x\ge \delta>1$, then we have

$$\zeta'(x)=-\sum_{n=1}^\infty \frac{\log(n)}{n^x}$$

for all $x>1$. And we are done!

Answer 2

Here's another way that can be used if you know the theory of complex variables.

Let $C$ be a counterclockise-oriented circle in $\mathbb C$ within the half-plane $\operatorname{Re} z>1.$ Then

\begin{align} & \int_C \sum_{n=1}^\infty \frac 1 {n^z} \, dz \\[8pt] = {} & \sum_{n=1}^\infty \int_C \frac{dz}{n^z} \text{ by Fubini's theorem, which can} \\ & \text{be applied since } z\mapsto\frac 1 {n^z} \text{ is continuous on the} \\ & \text{compact set }C \\[8pt] = {} & \sum_{n=1}^\infty 0 = 0. \end{align} The conclusion then follows by Morera's theorem.