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.
Show that the series $\sum_{n=1}^\infty \sin( \frac{x}{n^2}) $ converges pointwise but not uniformly
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1$\begingroup$ 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 $x$. $\endgroup$– Minus One-TwelfthMay 22, 2020 at 22:18
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$\begingroup$ 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? $\endgroup$– MathiasMay 22, 2020 at 22:21
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$\begingroup$ 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). $\endgroup$– Minus One-TwelfthMay 22, 2020 at 22:25
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$\begingroup$ 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) $\lim\limits_{\theta\to0}\frac{\sin \theta}{\theta}=1$.) $\endgroup$– Minus One-TwelfthMay 22, 2020 at 22:27
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1$\begingroup$ If $\sum a_n(x)$ converges uniformly, then $a_n(x)\to 0$ uniformly. Clearly $\sin(x/n^2)$ fails to converge uniformly to $0$. $\endgroup$– Mark ViolaMay 22, 2020 at 22:59
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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 $M>N$: $$ \begin{aligned} \|S_M-S_N\| &=\max_{-K\le x\le K}\left|\sum_{N<n\le M}\sin\frac x{n^2}\right| \\ &\le \max_{-K\le x\le K}\sum_{N<n\le M}\left|\sin\frac x{n^2}\right| \\ &\le \max_{-K\le x\le K}\sum_{N<n\le M}\left|\frac x{n^2}\right| \\ &\le K\sum_{N<n\le M}\frac 1{n^2} \\ &\le K\sum_{N<n\le M}\frac 1{n(n-1)} \\ &\le \frac KN\ . \end{aligned} $$ 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.
