Show $\displaystyle\sum_{n=1}^\infty \frac{\mathrm{ln}(n)}{n^x}$ is not uniformly convergent on $(1,\infty)$.
I thought I can show that $\displaystyle\sup_{x\in (1,\infty)}\left|\frac{\mathrm{ln}(n)}{n^x}\right|\not\to 0$, but it does, so this doesn't help.
Use the fact that the series diverges at $x = 1$. Rewrite the series as
$$f(x) := \sum_{n = 1}^{\infty} \frac{\ln n}{e^{x \ln n}}$$
So the error term can be expressed as
$$\left|f(x) - \sum_{n = 1}^{N - 1} \frac{\ln n}{e^{x \ln n}}\right| = \sum_{n = N}^{\infty} \frac{\ln n}{e^{x \ln n}} > \frac{\ln N}{e^{x \ln N}}$$
Now if $x \approx 1$, the denominator is very small, while the numerator is very large; for example, if $N$ is chosen so that $\ln N = 1 / x$, the error is $\ln N$. You can use this to finish the proof.