I've stumbled upon a particularly unpleasant series, and I can't quite seem to crack it.
$\sum\limits_{n=1}^{\infty}\dfrac{\ln(1+nx)}{n^{2}} $
I need to show uniform convergence on any interval of the form [0,K], where $0<K<\infty$. As a hint for the series, I have been told that the series $\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{3/2}} $ is convergent. I'm assuming I need to use a M-test, and therefore prove that this second series is larger than the first, but I haven't been able to do that as of yet.
We have the following inequality for non-negative $t$: $$\ln(1+t) \leq \sqrt{t} .$$ Thus for $x \in [0,K]$ we have $$\left\vert \frac{\ln(1+nx)}{n^2} \right\vert \leq \frac{\sqrt{nx}}{n^2} =\frac{\sqrt{x}}{n^{3/2}}\leq \frac{\sqrt{K}}{n^{3/2}} .$$ Conclude using the Weierstraß M-test.
Edit: In order to prove the inequality, consider $f(t)=\sqrt{t}-\ln(1+t)$ on the interval $[0,\infty)$. This function vanishes at $t=0$, and has derivative $$f'(t)=\frac{1}{2\sqrt{t}}-\frac{1}{1+t}=\frac{1+t-2\sqrt{t}}{2\sqrt{t}(1+t)}=\frac{(1-\sqrt{t})^2}{2\sqrt{t}(1+t)} \geq 0 ,$$ so that it is non-decreasing.