I have to prove that the following function has all higher order derivatives. $$ f(x)=\sum_{n=1}^{\infty}u_n(x),\ \ u_n(x)=\frac{1}{n^x},\ \ E=(1,+\infty) $$
Here is what I did:
$$
\left(\frac{1}{n^x}\right)^{(k)}=(-1)^k\ln^kn\cdot\frac{1}{n^x}
$$
Therefore, each derivative is continuous on $E$.
Also, I showed that $\sum_{n=1}^{\infty}\frac{1}{n^x}$ is convergent at some point $x=2\in E$.
The last thing that I have not proved so far is showing that
$\sum_{n=1}^{\infty}(u_n(x))^{(k)}=\sum_{n=1}^{\infty}(-1)^k\ln^kn\cdot\frac{1}{n^x}$ is uniformly convergent on $E$.
I tried to bound $\left(\frac{1}{n^x}\right)^{(k)}\leqslant\frac{1}{n^{x-k}}$ and show that $\sum_{n=1}^{\infty}\frac{1}{n^{x-k}}$ is convergent for $x$ large enough but then I realized that the previous inequality must be met for any $x\in E$. So, I am a bit confused right now.
It is false that it is uniformly convergent on $E$.
Nevertheless, you can prove that it converges normally on every $[a,+\infty)$ with $a > 1$. For instance, for $k = 0$, $$ \forall x \in [a,+\infty),\quad \frac{1}{n^x} \leq \frac{1}{n^a}$$ and $\sum_n 1/n^a$ converges.
It gives that $f$ is continuous on every $[a,+\infty)$, so $f$ is continuous on $E$.
You can do the same thing for all $k \geq 1$.