<s>[INST] Please show that $f(x)=\sum \limits_{n=1}^\infty \frac{x^2}{x^2+n^2}$ is continuous on $\Bbb R$ 
Let $f(x)=\sum \limits_{n=1}^\infty \frac{x^2}{x^2+n^2}$ on $\Bbb R$. Please show that $f$ is continuous on $\Bbb R$.

I thought if I can prove that $f$ converges uniformly on $R$, then we are done. But I found that Show that $\sum_{n=1}^{\infty}\frac{x^{2}}{x^{2}+n^{2}}$ does not converge uniformly on $(-\infty,\infty)$., we have to find some other ways to prove it.
From Is $f(x)=\sum_{n=1}^\infty\frac{nx^2}{n^3+x^3}$ uniformly continuous on $[0,\infty)$? and Show $ f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$ ,$\ g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + x^3)^2}$ are bounded on $[0, \infty)$., I think I can use the same method to prove this question.
I think $f$ converges uniformly on any bounded intervals, hence $$f'(x)=2x\sum \limits_{n=1}^\infty \frac{1}{n^2+x^2}+x^2\sum \limits_{n=1}^\infty \frac{-2x}{(n^2+x^2)^2}$$
Let $g(x)=\sum \limits_{n=1}^\infty \frac{x}{n^2+x^2}$ and $h(x)=\sum \limits_{n=1}^\infty \frac{x^3}{(n^2+x^2)^2}$ on $\Bbb R$. We need to show that $g$ and $h$ are bounded on $\Bbb R$.
First, we discuss $g$.
For $x\in [-1,1]$, $$\left|\frac{x}{n^2+x^2} \right| \le \frac{1}{n^2}$$, so $\left|g(x) \right|\le \sum \limits_{n=1}^\infty \left|\frac{x}{n^2+x^2} \right|\le \sum \limits_{n=1}^\infty \frac{1}{n^2} < \infty$.
For $|x|>1$ be fixed and $m\in \Bbb N$, for any $n\in \Bbb N$ such that $$x(m-1)\le n < mx$$, we have $$\left|\frac{x}{n^2+x^2} \right| \le \left|\frac{x}{x^2(m-1)^2+x^2}\right|=\left|\frac{1}{x}\frac{1}{(m-1)^2+1}\right|$$ and there are at most $\lfloor x \rfloor +1$ of such $n$. $$\left|g(x) \right|\le \sum \limits_{n=1}^\infty \left|\frac{x}{n^2+x^2} \right|\le \sum \limits_{n=1}^\infty \left|\frac{\lfloor x \rfloor +1}{x}\right|\left|\frac{1}{(m-1)^2+1}\right|\le \sum \limits_{n=1}^\infty 2\frac{1}{(m-1)^2+1} < \infty  $$
Hence, $g$ is bounded.
Now, consider $h$, if $x\in [-1,1]$, then $$\left|\frac{x^3}{(n^2+x^2)^2} \right|\le \frac{1}{n^4}$$
For $|x|>1$ and $m\in \Bbb N$ be fixed, for any $n\in \Bbb N$ such that  $$x(m-1)\le n < mx$$, we have $$\left|\frac{x^3}{(n^2+x^2)^2} \right| \le \left|\frac{x^3}{(x^2+(m-1)^2x^2)^2} \right|=\left|\frac{1}{x}\frac{1}{(1+(m-1)^2)^2} \right|$$ and there are at most $\lfloor x \rfloor +1$ of such $n$.
$$\left|h(x) \right|\le \sum \limits_{n=1}^\infty \left|\frac{x^3}{(n^2+x^2)^2} \right|\le \sum \limits_{n=1}^\infty \left|\frac{\lfloor x \rfloor +1}{x}\right|\left|\frac{1}{((m-1)^2+1)^2}\right|\le \sum \limits_{n=1}^\infty 2\frac{1}{((m-1)^2+1)^2} < \infty$$ Hence, $h$ is bounded.
Finally, by mean value theorem, we can prove that $f$ is uniformly continuous on $\Bbb R$.
Appreciate any suggestion.
 [/INST] A much shorter proof: continuity is local property. To prove that $f$ is continuous it is enough to prove that it is  continuous at each point. Take any $x \in \mathbb R$. In $[x-1,x+1]$ the series converges uniformly by M-test since it is dominated by a constant times $\sum \frac  1 {n^{2}}$. Hence $f$ is continuous on $[x-1,x+1]$ and we are done.
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