# Show that $\sum\limits_{n=1}^{\infty}\frac{x}{1+n^2x^2}$ is not uniformly convergent in $[0,1]$

Show that $\sum\limits_{n=1}^{\infty}\frac{x}{1+n^2x^2}$ is not uniformly convergent in $[0,1]$.

I was thinking in the direction of taking the maximum value of each term $\frac{x}{1+n^2x^2}$, which is $\frac{1}{2n}$, and of summing them. That is clearly a divergent series.

But then those maximum values don't occur at the same value of $x$ for each term. For the $n$th term the maximum occurs at $x = \frac{1}{n}$.

So, how to proceed?

You are on the right track. For clarity, let's denote the partial sum $\sum_{n = 1}^N \frac{x}{1 + n^2x^2}$ by $S_N(x)$. One way to show the result is to investigate $\sup_{x \in [0, 1]}|S_{2N} (x) - S_{N}(x)|$: \begin{align} \sup_{x \in [0, 1]} |S_{2N}(x) - S_N(x)| & = \sup_{x \in [0, 1]}\sum_{n = N + 1}^{2N} \frac{x}{1 + n^2x^2} \geq \sum_{n = N + 1}^{2N} \frac{1/N}{1 + n^2/N^2} \\ & \geq \frac{1}{N} \times N \times \frac{1}{1 + (2N)^2/N^2} = \frac{1}{5} \end{align} which doesn't converge to $0$ as $N \to \infty$. Thus we do not have uniform convergence (if $\{S_N(x)\}$ converges uniformly, then the above quantity is bounded to converge to $0$ as $N \to \infty$, in view of Cauchy's criterion).

• How did you get $\frac{x}{1+n^2x^2} \ge \frac{1/N}{1+n^2/N^2}$? May 20, 2016 at 19:15
• Since $1/N \in [0, 1]$ and it's supremum. Don't look it termwisely but treat it as a whole function. May 20, 2016 at 19:16
• $1/2N$ is the supremum of $\frac{x}{1 + N^2x^2}$, not of the every term of the summation. Also, even if it were supremum of every term, the inequality sign should be the other way around. May 20, 2016 at 19:20
• Sorry for the confusion. I got it now. May 20, 2016 at 19:21
• @Prince Kumar Glad to know that. May 20, 2016 at 19:22

The problem is that $$\lim_{x \rightarrow 0}\; \sum_{n=1}^{\infty} \frac{x}{1 + n^2 x^2}$$ is not $0$, so the function defined by the series isn't continuous. You can see that by evaluating at $x = 1/k:$ $$\sum_{n=1}^{\infty} \frac{1/k}{1 + n^2 / k^2}$$ is a Riemann sum for the integral $$\int_0^{\infty} \frac{1}{1+y^2} \, \mathrm{d}y$$ and so it tends to $\pi/2$ as $k \rightarrow \infty.$

• You are using some fact about improper Riemann integrals. Which one is it?
– zhw.
May 20, 2016 at 23:31
• @zhw. Yes, technically you would need to compute Riemann sums for finite integrals $\int_0^N \frac{1}{1+y^2} \, \mathrm{d}y$ for large $N$. Since $\frac{1}{1+y^2}$ is positive and monotone there will be no problem swapping limits in $N$ and $k$. For stranger functions like $\frac{\sin(x)}{x}$ you would have to be more careful. May 20, 2016 at 23:58