Is $f(x)=\sum_{n=1}^{\infty}\frac{x}{1+n^2x^2}$, $x\in[0,1]$ continuous on $[0,1]$ Let $f(x)=\sum_{n=1}^{\infty}\frac{x}{1+n^2x^2}$, $x\in[0,1]$.
Question: Show that $f$ is Lebesgue integrable and determine whether $f$ is continuous on $[0,1]$.
For the first part, I have no problem, I showed that it is Lebesgue integrable. But for the second part, I could not reach any result by using the definition. Also, I'm not sure if I should use the first part.  Can you give me a hint for this part?
 A: Let $x_n = 1/n$ and note that
$$f(x_n) \geqslant \sum_{k=1}^n \frac{x_n}{1+k^2 x_n^2} \geqslant \frac{nx_n}{1+n^2x_n^2} = \frac{1}{2}$$
Since $x_n \to 0$ and $f(x_n) \not\to 0$, we have a discontinuity at $x = 0$.  Elsewhere the function is continuous since convergence of the series is uniform on $[a,1]$ for all $0<a < 1$.

A clue that there is a discontinuity at $x=0$ is that the series converges uniformly on $[a,1]$ for all $0 < a <1$ but not on $[0,1]$.  Uniform convergence is, of course, only a sufficient condition for continuity.
A: I thought it might be instructive to present an approach in which we establish directly that $f(x)$, as given by the series
$$f(x)=\sum_{n=1}^\infty\frac{x}{1+n^2x^2}\tag1$$
is discontinuous at $x=0$.  We do so first by evaluating $\lim_{x\to0^+}f(x)$ and finding that the limit is not $0$.
We follow that analysis by deriving a closed form expression for $f(x)$ to show that $f(x)$ is discontinuous.  We now proceed with these analyses.


Direct Evaluation of the Limit

For $0<x$, the general terms of the series in $(1)$ monotonically decrease for all $n\ge 1$.  Hence, we can assert that for $0<x$
$$\begin{align}
\underbrace{\int_1^{N+1}\frac{x}{1+t^2x^2}\,dt}_{\arctan((N+1)x)-\arctan(x)}\le \sum_{n=1}^N \frac{x}{1+n^2x^2} \le \underbrace{\int_0^N \frac{x}{1+t^2x^2}\,dt}_{\arctan(Nx)}\tag2
\end{align}$$

Letting $N\to \infty$ in $(2)$, we obtain
$$\frac{\pi}{2}-\arctan(x)\le f(x) \le \frac{\pi}{2}\tag3$$
whereupon letting $x\to 0^+$, applying the squeeze theorem to $(3)$, and exploiting the oddness of $f(x)$  yields
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0^{\pm}}f(x)=\pm\frac\pi2}$$

Inasmuch as $f(0)=0\ne \lim_{x\to 0^{\pm}}f(x)$, $f(x)$ is discontinuous from both the left and right at $0$.  And we are done!



Deriving a Closed-Form Solution

Using contour integration, we can evaluate $f(x)$ in closed form.  Let $g(z)=\frac{(\cot(\pi z)}{z^2+|a|^2}$.  Then, we have
$$\lim_{N\to \infty}\oint_{|z|=N+1/2}g(z)\,dz=0$$
And from the residue theorem, we have for $|a|<N$
$$\oint_{|z|=N+1/2}g(z)\,dz=2\pi i \left(\frac1\pi\sum_{n=-N}^N \frac{1}{n^2+|a|^2}-\frac{\coth(\pi |a|)}{|a|}\right)\tag4$$
Letting $N\to\infty$ in $(4)$, we find that
$$\sum_{n=1}^\infty \frac{1}{n^2+|a|^2}=\frac\pi2 \left(\frac{\coth(\pi|a|)}{|a|}-\frac{1}{\pi |a|^2}\right)\tag5$$
Let $a=1/x$ and multiply by $1/x$ to find
$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{x}{1+n^2x^2}=\frac\pi2 \text{sgn}(x)\coth(\pi/|x|)-\frac x2}\tag6$$
Letting $x\to 0^{\pm}$ in $(6)$ and we find that
$$\lim_{x\to0^{\pm}}f(x)=\pm \frac\pi2$$
as expected!
