Does $\sum_{n=1}^{\infty}{\ln{n}\over n^x}$ converge uniformly?

Question

Suppose we have $\zeta(x)=\sum_{n=1}^{\infty}{1\over n^x}, x\in(1,\infty)$. I would like to show that $\zeta'(x)$ is continuous in $(1,\infty)$. Let $f_n(x)=n^{-x}$, then $f_n'(x)={\ln{n}\over n^x}$. It would suffice to show that $\sum_{n=1}^{\infty}{\ln{n}\over n^x}$ is uniformly convergent. It seems to me that this is easily done with the Weierstrass M-test. However, a reputable source has suggested that $\sum_{n=1}^{\infty}{\ln{n}\over n^x}$ is not uniformly convergent but rather almost uniformly convergent. Am I missing some important assumptions?

EDIT:

The Weierstrass M-test can only be applied if the inequality between the series holds for $x \in (1,\infty)$, so $\sum_{n=1}^{\infty}{\ln{n}\over n^x}$ is not an appropriate series for comparison (the comments below show that there isn't one).

Answer 1

hint

To prove that your function is continuous at $(1,+\infty) $, you prove it is continuous at $[a,+\infty) $ for given $a>1$.

so you need to prove uniform convergence at $[a,+\infty) $ by normal convergence.

$$\frac {1}{n^x}\leq \frac {1}{n^a} $$

with $a>1$ (use Bertrand series).

Answer 2

To show that $\zeta'$ is continuous in $(1, +\infty)$ it is not necessary to prove that $\sum f_n'$ is uniformly convergent in $(1,+\infty)$ (indeed, this is false).

Since continuity is a local notion, it is enough to prove that $\sum f_n'$ converges uniformly on $[a, +\infty)$ for every $a > 1$.