Uniform convergence of $\sum_{n=1}^\infty \frac{1}{n^3+x}$

Question

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 $-1<x<0$.

Answer 1

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.

Answer 2

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 $r\to -1$.

Answer 3

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.