Differentiability of the Riemann function 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 $g(x) := \frac{1}{n^x}$
 A: 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!
A: 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.
