# Characterizing where the series $f(x)=\sum\limits_{n=1}^{\infty}\frac{1}{1+n^2|x|}$ is uniformly convergent.

Let $f(x)=\sum\limits_{n=1}^{\infty}\frac{1}{1+n^2|x|}$.

1. For what values of x does this series converge.
2. Characterize the subsets of $\mathbb{R}$ on which the series converges uniformly.
1. There are only two cases to deal with $x=0$ and $x>0$. If $x<0$, because of the absolute value, we just have the $x>0$ case again. For $x>0$ we have $$\sum\limits_{n=1}^{\infty}\frac{1}{1+n^2|x|}\leq \sum\limits_{n=1}^{\infty}\frac{1}{n^2|x|}=\frac{1}{|x|}\sum\limits_{n=1}^{\infty}\frac{1}{n^2}<\infty.$$ For $x=0$ we have a divergent series.

2. On any interval that contains $0$ convergence will not occur. On any inverval that does not contain $0$ there is uniform convergence by the Weierstrass M-Test.

If we have an open interval where $0$ is an endpoint, then we don't have uniform convergence. See,

$$\left|\sum\limits_{n=1}^{\infty}\frac{1}{1+n^2|x|}\right|<\frac{1}{2}$$

and let $x=\pm \frac{1}{n^2}$.

In (2), "any inverval that does not contain 0" should read "any interval at positive distance from $0$" (consider the interval $(0,1)$ for a counterexample).
To disprove the uniform convergence, one should not consider the sum $\sum\limits_{n=1}^{+\infty}$ (as you seem to be doing) but sums $\sum\limits_{n=N}^{+\infty}$. Then your idea to consider $x=1/N^2$ is a good one since $u_N(1/N^2)=1/2$, with $u_n(x)=1/(1+n^2|x|)$. Hence, for every $N$, $\sum\limits_{n=N}^{+\infty}u_n(1/N)\gt u_N(1/N)=1/2$ and the rest $\sum\limits_{n=N}^{+\infty}u_n$ of the series does not converge to zero uniformly on any interval whose closure contains $0$.