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 $-1<x<0$.
 A: Since $\forall n\geq 2$, $\max_{x\in (-1,1)} |f_n(x)| \leq \frac{1}{n^3-1}$ and $\sum_{n=2}^\infty \frac{1}{n^3-1}$ converges, your series has normal convergence and thus converges uniformly on $(-1,1)$.
Now, let $(g_n)_{n\in \mathbb{N}}$ be a sequence of functions that converges uniformly towards $g$ on a given interval $I$. Let $h$ be a well-defined function over $I$. Since $(g_n)_{n\in \mathbb{N}}$ converges uniformly, we have:
$$
\forall \epsilon >0, \exists N_\epsilon\in\mathbb{N};\forall n\in \mathbb{N},n\geq N_\epsilon, \forall x\in I, |g_n(x)-g(x)|\leq \epsilon
$$
Then:
$$
\forall \epsilon >0, \exists N_\epsilon\in\mathbb{N};\forall n\in \mathbb{N},n\geq N_\epsilon, \forall x\in I, |(g_n(x)+h(x))-(g(x)+h(x))|\leq \epsilon
$$
So $(g_n + h)_{n\in \mathbb{N}}$ converges uniformly to $g+h$ over $I$.
Then, let $I=(-1,1)$, $g_n =\sum_{k=2}^n f_n$ be your series and $h=f_1$ be the missing part and you may conclude.
A: 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 $r\to -1$.
A: Actually, you could use a simple comparison test.  Since $|\frac 1{n^2}|>|\frac 1 {n^3-x}| for -1<x<1$, and since the sum of the reciprocal squares is uniformly convergent, then it follows that your original sum is also convergent.
