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This series \begin{equation} \sum_{n=1}^{\infty} \frac{\sin(n)}{n^2} \end{equation} converges, because: $\sin(n)\leq |\sin(n)|\leq 1$ and the series \begin{equation} \sum_{n=1}^{\infty} \frac{1}{n^2} \end{equation} converges.

Now, I have a curiosity. Let $k=k(n)$ the sequence that minimizes the distance $|\pi k -n|\leq \frac{\pi}{2}$. Then: \begin{equation} |\sin(n)|=\sin|\pi k -n| \end{equation} Also, from $|\pi k -n|\leq \frac{\pi}{2}$, we have \begin{equation} \left|\pi -\frac{n}{k(n)}\right|\leq \frac{\pi}{2|k(n)|} \end{equation} Let $n=n(k)$, rewriting the above relation: \begin{equation} \left|\pi -\frac{n(k)}{k}\right|\leq \frac{\pi}{2|k|} \end{equation} If k tends to infinity, n also diverges. Then we can write that: $n/k\approx \pi$?

Furthermore: \begin{equation} |\sin(n)|=\sin|\pi k -n|\leq|\pi k -n| \end{equation} so: \begin{equation} \sum_{n=1}^{\infty} \frac{|\pi k -n|}{n^2}=\sum_{n=1}^{\infty} k \frac{|\pi -n/k|}{n^2} \end{equation} Can I change the index from n to k? I.e. \begin{equation} k \frac{|\pi -n/k|}{n^2}\approx k \frac{|\pi -n/k|}{\pi^2 k^2} \ \ \ \ (n/k\approx \pi) \end{equation} and so the series: \begin{equation} \sum_{n=1}^{\infty} \frac{|\pi k -n|}{n^2} \end{equation} has the same behavior of \begin{equation} \sum_{k=1}^{\infty} \frac{|\pi -n/k|}{k} \end{equation} That is, both series converge or both diverge. This reasoning is correct? We welcome suggestions. Thank you very much.

Mark
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    While not answering your actual approach and the resulting question (and unless I'm missing something), we have $|\pi - n(k)/k|<\frac{1}{k}$, since $n$ was chosen to minimize the absolute value of the difference. Thus, we also have $\sum_{k=1}^\infty \frac{|\pi - n(k)/k|}{k}<\sum_{k=1}^\infty \frac{1}{k^2}$ which converges. – Peter Košinár Nov 27 '13 at 08:45
  • Have you looked at Convergence of $\displaystyle \sum_{n=1}^\infty\frac1{n^3\sin^2 n}$ on MathOverflow? The response involves connections to Diophantine approximation which will be necessary for your discussion of convergence. – cactus314 Nov 30 '13 at 21:21
  • @johnmangual. Thanks, I just read it. Maybe I did not understand the question, but I think my case is different. The author of the response you quoted, says: "Neglecting the terms of the sum for which $n|\sin n|\ge n^\varepsilon$ as they all contribute only to the `convergent part' of the sum (...)" and thanks to this appointment, he can write "the question is equivalent to the one for the series: $\sum_{n:n|\sin n|< n^\varepsilon}\frac1{n^3\sin^2n}$" and $\sin|\pi q-n|=|\sin n|< \frac1{n^{1-\varepsilon}}$. In my exercise this is not possible. Or am I wrong? – Mark Dec 01 '13 at 10:46

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Write explicitly $n = k\pi + \epsilon$ with $0 < \epsilon < \pi$. Then

$$ \sum_{n=1}^\infty \frac{|\pi k - n|}{n^2} = \sum_{n=1}^\infty \frac{|\epsilon|}{(k\pi + \epsilon)^2}< \sum_{n=1}^\infty \frac{|\epsilon|}{(k\pi )^2} < \pi \sum_{n=1}^\infty \frac{1}{(k\pi )^2} = \frac{\pi}{6}$$

While it is always true that $ |\sin(n)|=\sin|\pi k -n|\leq|\pi k -n| $, we don't always know that $|\pi k - n|$ is small.

By Dirichlet's approximation theorem or just the pigeonhole principle we can find $n$ for which $$\min_{k \in \mathbb{Z}} | \pi k - n| \approx 0 $$

However, this same logic show we can find infinitely many $n$ for which $$\min_{k \in \mathbb{Z}} | \pi k - n| \approx \frac{\pi}{17} $$

Or else we get a contradiction of the pigeonhole principle.

In fact, we can show the integers $\mathbb{Z}$ will be equidistributed mod $\pi$, that the sequence $\frac{n}{\pi}$ hits all the values of $[0,1]$ "equally often".

cactus314
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  • I'm sorry, but I can not understand your reasoning. I do not understand what is the connection with Fourier series $\sum_{n=1}^\infty \sin (2\pi n x)$, I do not understand the usefulness of this passage: $\sum_{n=1}^{\infty} \frac{\sin(n)}{n^2} = \sum_{k=-M}^{M-1}\sum_{ \frac{k}{M} < \sin n < \frac{k+1}{M} } \frac{ \sin n}{n^2} \approx \sum_{k=1}^{M-1} \frac{k}{M}\sum_{ \frac{k}{M} < \sin n < \frac{k+1}{M} } \frac{ 1}{n^2}$ From this passage how do you deduce the convergence of the series? ... Maybe I do not understand because of my bad English – Mark Dec 01 '13 at 19:49
  • The 1st part is the most important. Your estimate \begin{equation} |\sin(n)|=\sin|\pi k -n|\leq|\pi k -n| \end{equation} is not correct. Then I tried to find an explicit value for \begin{equation} \sum_{n=1}^{\infty} \frac{\sin(n)}{n^2} \end{equation} it is related to the dilogarithm function. – cactus314 Dec 01 '13 at 19:56
  • Why my estimate isn't correct? I know that $|\sin t|\leq |t|$ – Mark Dec 01 '13 at 20:48
  • @johnmangual: I think the actual question is in the end where he compares the two series. He explicitly says that he chooses $k(n)$ such that $n/k(n) \to \pi$. – Beni Bogosel Dec 02 '13 at 16:54