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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?

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2 Answers 2

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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).

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  • $\begingroup$ How did you get $ \frac{x}{1+n^2x^2} \ge \frac{1/N}{1+n^2/N^2} $? $\endgroup$ May 20, 2016 at 19:15
  • $\begingroup$ Since $1/N \in [0, 1]$ and it's supremum. Don't look it termwisely but treat it as a whole function. $\endgroup$
    – Zhanxiong
    May 20, 2016 at 19:16
  • $\begingroup$ $1/2N$ 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. $\endgroup$ May 20, 2016 at 19:20
  • $\begingroup$ Sorry for the confusion. I got it now. $\endgroup$ May 20, 2016 at 19:21
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    $\begingroup$ @Prince Kumar Glad to know that. $\endgroup$
    – Zhanxiong
    May 20, 2016 at 19:22
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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 $k \rightarrow \infty.$

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    $\begingroup$ You are using some fact about improper Riemann integrals. Which one is it? $\endgroup$
    – zhw.
    May 20, 2016 at 23:31
  • $\begingroup$ @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. $\endgroup$
    – user341050
    May 20, 2016 at 23:58

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