TITLE: proof that $f(x)=\sum_{n=1}^\infty \frac{\ln(x+n)}{x^2 + n^2}$ converges uniformily
QUESTION [6 upvotes]: Prove that $f(x)=\sum_{n=1}^\infty \frac{\ln(x+n)}{x^2 + n^2}$  converges uniformly for $x\geq 0$.
This exercise is in a text about uniform convergence, I've tried to use Weierstrass test with no success. trying to quote with
$a_n=\sqrt{x+n}/(x^2 + n^2)$
so that
$\ln(x+n)/(x^2 + n^2)\leq\sqrt{x+n}/(x^2 + n^2)\,,$
but then I can't find a criteria to prove that  $\sqrt{x+n}/(x^2 + n^2)$ converges (and if I prove this then $f(x)$ converges uniformly as Weierstrass test claims)

REPLY [2 votes]: Hint: $(x+n)^2 \le 2(x^2 + n^2)$.

 $$\dfrac{\sqrt{x+n}}{x^2+n^2} \le \dfrac{\sqrt[4]{2}\sqrt[4]{x^2+n^2}}{x^2+n^2} = \dfrac{\sqrt[4]{2}}{(x^2+n^2)^{3/4}} \le \dfrac{\sqrt[4]{2}}{n^{3/2}}.$$