a question about uniform convergence of a series The problem is to investigate the uniform convergence of $\sum\limits_{n=0}^\infty \frac{x^2}{x^2+n^2}.$  Any help will be appreciated.
Thanks in advance.
 A: Hints:
Let $f_n(x)={x^2\over x^2+n^2}$.
1) On the interval $I=[-a,a]$, we have $|f_n(x)|\le {a^2\over n^2}$ for all $n$ and $x\in I$. How could you then deduce that $\sum\limits_{n=1}^\infty f_n(x)$ converges uniformly on $I$? One can generalize here to the case where $I$ is any bounded set.
2) Note that the sequence $(f_n)$ does not converge uniformly to $0$ on $(-\infty,\infty)$. Could $\sum\limits_{n=1}^\infty f_n(x)$ then converge uniformly on $(-\infty,\infty)$?

Recall that a series $\sum\limits_{n=1}^\infty g_n(x)$ converges uniformly on a set $A$ if and only if it is uniformly Cauchy on $A$; that is:
For every $\epsilon>0$, there is an $N$ so that for all $m\ge k\ge N$, we have
$$
\biggl|\,\sum_{n=k}^m g_n(x)\,\biggl|<\epsilon
$$
for all $x\in A$. (Of course, for part 1), you may recall a certain German's $M$-test.)
A: On any interval $[a, b]$, let $M = \sup\{|x| : x \in [a, b]\}$. We have:
$$
\frac{x^2}{x^2 + n^2} \le \frac{M}{n^2}
$$
Since $\sum_{n=1}^\infty M/n^2 < \infty$, by the Weierstrass M-test, the series converges uniformly on $[a, b]$.
A: As a bonus, this series not only converges but there is a closed form expression for the sum. Using the method presented here, we study the integral of $$ f(z) = \frac{x^2}{x^2+z^2} \pi \cot(\pi z)$$ along a circle of radius $R$ in the complex plane where $R$ goes to infinity. This integral vanishes in the limit, so the residues at the poles of $f(z)$ must sum to zero. Let $$S = \sum_{n\ge 1} \frac{x^2}{x^2+n^2} .$$ We thus have $$ \operatorname{Res}_{z=-ix} f(z) + \operatorname{Res}_{z=+ix} f(z) + \operatorname{Res}_{z=0} f(z) + 2S = 0,$$ where we have used the fact that $$\operatorname{Res}_{z=m} \pi\cot(\pi z) = 1,$$ with $m$ an integer. 
This gives $$\frac{x^2}{-2ix} \pi\cot(-\pi ix) + \frac{x^2}{+2ix} \pi\cot(+\pi ix) + 1 + 2S=0$$ or $$ \frac{x^2}{2ix} \pi\cot(\pi ix) + \frac{x^2}{2ix} \pi\cot(+\pi ix) + 1 + 2S=0$$ or $$ \frac{x}{i} \pi\cot(\pi ix) + 1 + 2S=0$$ and finally $$ S = -\frac{1}{2} + \frac{\pi}{2}x\coth(\pi x).$$ It follows that for the original sum with the term at $n=0$ we have $$\sum_{n\ge 0} \frac{x^2}{x^2+n^2} = \frac{1}{2} + \frac{\pi}{2}x\coth(\pi x).$$
A: $\sum \dfrac{x^2}{x^2+n^2}\le x^2\sum \frac{1}{n^2}$
