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?

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.
