# Show that the series $\sum_{n=1}^\infty \sin( \frac{x}{n^2})$ converges pointwise but not uniformly

For $$x \in \mathbb{R}$$ consider the series defined as $$S = \sum_{n=1}^\infty \sin( \frac{x}{n^2})$$ I then have to show that $$S$$ converges pointwise but not uniformly. I know that for all $$x \in \mathbb{R}$$ and for all $$n \in \mathbb{N}$$ that $$|\sin ( \frac{x}{n^2})| \leq \frac{|x|}{n^2}$$ and I then thought I could use the fact that $$\sum_{n=1}^\infty \frac{1}{n^2}$$ is a convergent series. Thus Weiterstrass' M-test would say that $$S$$ converges both uniformly and absolutely but as I have to show that $$S$$ does not converge uniformly I am little bit lost. How does $$x$$ change the fact that we cannot use Weiterstrass' M-test in this way? And how do I then show pointwise convergence? Thanks for your help.

• In the Weierstrass $M$-test, the bounds you use for the terms of the series are not allowed to depend on $x$ (but may depend on $n$). So you can't use $\frac{|x|}{n^2}$ like you did here, since that depends on $x$. May 22, 2020 at 22:18
• Thanks. I find it pretty easy to show that a series is uniformly convergent if I am allowed to use Weiterstrass M-test but if I am not I am pretty lost. What can I do here to show pointwise convergence? May 22, 2020 at 22:21
• In this question, it is asking you to show that it is not uniformly convergent, so Weierstrass $M$-test won't help here (that only lets you show that a series is uniformly convergent). May 22, 2020 at 22:25
• First of all, try to show pointwise convergence, and to do this, you just need to show that for any fixed $x\in\Bbb{R}$, the series $\sum\limits_{n=1}^{\infty} \sin\left(\frac{x}{n^2}\right)$ is a convergent series. Try and think back to your study of series convergence to think about how to show this. (Hints: Remember the rules 1) if $\sum |a_n|$ converges, then $\sum a_n$ converges, 2) the Limit Comparison Test, 3) $\lim\limits_{\theta\to0}\frac{\sin \theta}{\theta}=1$.) May 22, 2020 at 22:27
• If $\sum a_n(x)$ converges uniformly, then $a_n(x)\to 0$ uniformly. Clearly $\sin(x/n^2)$ fails to converge uniformly to $0$. May 22, 2020 at 22:59

Let us make things more detailed. Let $$S_N$$ be the function $$\Bbb R\to \Bbb R$$, given by the partial sum $$S_N(x) =\sum_{1\le n\le N}\sin\frac x{n^2}\ .$$ The pointwise convergence, and in the same time the convergence on each compact set $$I(K)=[-K, K]$$ follows with the argument from the OP. We have on such an interval w.r.t. the supremum = max norm $$\|\cdot\|=\|\cdot\|_\infty$$ the estimation for two indices $$M,N$$ with $$M>N$$: \begin{aligned} \|S_M-S_N\| &=\max_{-K\le x\le K}\left|\sum_{N So given an $$\epsilon>0$$ we make the choice of $$N(\epsilon):=\frac K\epsilon$$, for for all $$M,N$$ with $$M>N>N(\epsilon)$$ we have $$\|S_M-S_N\|<\epsilon$$. The space of continuous functions on the compact interval $$I(K)=[-K,K]$$ is a Banach space with the supremum norm, so there is a limit.
Let us show that there is no uniform convergence on $$\Bbb R$$. Assume the contrary. Then there exists a limit $$S$$, a continuous function. (Because it is continuous on each interval $$[-K,K]$$.) For $$\epsilon:=1/3$$ there exists an $$N(1/3)$$ so that for all $$N\ge N(1/3)$$ we have $$\|S-S_N\|\le 1/3$$. In particular, $$1=\|S_{N+1}-S_N\|\le 2/3$$. Contradiction.