Uniform convergence of $\sum_{n=1}^\infty \frac{1}{n^3+x}$ How do is show that for $-1<x<1$ the series $\sum_{n=1}^\infty \frac{1}{n^3+x}$ converges uniformly.
For $x\geq0$ I can make a convergent Majorant series, choosing $M_n = 1/n^3$, but I can't seem to determine a convergent Majorant series for $-1<x<0$.

Applying Weierstrass. For every $r \in (-1,1)$ you can consider $A_r = (r,1)$. Now apply M-test using that $\forall x \in A_r$
$$
\frac{1}{n^3+x} \leq \frac{1}{n^3+r}
$$
And obviously
$$
\sum_{n=1}^\infty \frac{1}{n^3+r} < \infty
$$
You can check it by comparation with $\sum_n 1/n^3$. So you prove uniform convergence in $A_r$. Now take limit $r\to -1$.
