I need to prove that the $\zeta$ function $$\zeta(x):=\sum^{\infty}_{k=1}\frac{1}{k^x}$$defines a function that has continuous derivatives of all orders.
Here's my attempt:
So, the $n$'th derivative is $$\frac{(-1)^n\ln^n(k)}{k^x}$$
So, we can see that this is continuous on $x\in(1,\infty)$ and that the partial sums $\zeta_s(x)=\sum^s_{k=1}\frac{(-1)^n\ln^n(k)}{k^x}$ are finite, so hence the function has continuous derivates of all orders.
Is this enough/correct?