# Does the series $\sum_{n=1}^{\infty}{\frac{\sin^2(\sqrt{n})}{n}}$ converge?

Does the following series converge? $$\sum_{n=1}^{\infty}{\frac{\sin^2(\sqrt{n})}{n}}$$

It shouldn't, but I have no idea how to prove it. I was wondering about Integral Criterion, but the assumptions are not satisfied. Or perhaps Dirichlet test would help, but then it should be shown that $\sum_{k=1}^n(\sin^2(\sqrt{k}))$ is bounded.

• $\sum(sin^2(\sqrt k))$ is not bounded. There exists a sub-sequence such thant every $k_n$ is nearly 1. – Doug M Mar 29 '16 at 16:56
• Use limit comparison test – user90533 Mar 29 '16 at 16:59
• I'm not sure if limit comparsion test helps. Function $\sin(x)$ can be compared with $x$ when $x$ tends to $0$. Here $x$ goes to $\infty$. – joseph Mar 29 '16 at 17:02
• This depends on the distribution of $\sin(n^{\frac12})$ within $[-1,1]$, but I can't find a relevant result right now. – dtldarek Mar 29 '16 at 17:22
• One idea: determine the ranges for which $\sin^2(\sqrt{n}) \geq \frac{1}{4}$. Then estimate the sum of the terms that fall in this category. This should correspond to roughly $\frac{(2k-1/3)^2\pi^2}{4} < n < \frac{(2k+1/3)^2\pi^2}{4}$ where $k$ is an integer. There are roughly $\propto k$ terms which should sum to $O(1/k)$ which diverges. – Winther Mar 29 '16 at 17:25

## 3 Answers

Observe that $\sin^2(\sqrt{n}) \geq 1/4$ iff $|\sin(\sqrt{n})| \geq 1/2$ iff $$\frac{\pi}{6} + k\pi \leq \sqrt{n} \leq \frac{5\pi}{6} + k\pi$$ for some nonnegative integer $k$. This chain of inequalities is equivalent to $$\left(\frac{\pi}{6} + k\pi\right)^2 \leq n \leq \left(\frac{5\pi}{6} + k\pi\right)^2$$ For a fixed $k$, the number of values of $n$ which satisfy the above is approximately $$\left(\frac{5\pi}{6} + k\pi\right)^2 - \left(\frac{\pi}{6} + k\pi\right)^2 = \frac{2\pi^2}{3} + \frac{4\pi^2}{3}k > 6+13k$$ Therefore, $$\sum_{n=1}^{\infty}\frac{\sin^2(\sqrt{n})}{n} > \sum_{k=1}^{\infty}\frac{6+13k}{4}\frac{1}{\left(\frac{5\pi}{6} + k\pi\right)^2}$$ which diverges by limit comparison with $\sum\frac{1}{k}$.

• Very nice. +1.${}$ – YoTengoUnLCD Mar 29 '16 at 18:17
• Excellent in its simplicity and directness. – DanielWainfleet Mar 29 '16 at 23:41

Denote $S_N=\sum_{n=1}^N\frac{\sin^2(\sqrt{n})}{n}$.

By the Euler-MacLaurin formula we have

$$S_N\sim_{\infty}\int_1^N dx\frac{\sin^2(\sqrt{x})}{x}+\mathcal{O}(1)$$

You can show this by observing that the derivates of $\frac{\sin^2(\sqrt{x})}{x}$ are $\mathcal{O}\left(\frac{1}{x^{1+m/2}}\right)$, where $m$ is the order of the derivative.

Performig a change of variables $x=y^2$ we get $$S_N\sim_{\infty}2\int_1^N dx\frac{\sin^2(y)}{y}+\mathcal{O}(1)\sim\log(N)+\mathcal{O}(1)$$

which shows that the sum is unbounded. The last asymptotic identity can be proved by using $\sin(x)^2=\frac{1}{2}(1-\cos(2x))$ combined with an integration by part.

• Just a small comment, your formula for $\sin^2x$ in the end has a sign problem. Except that, I like that you use asymptotics and the Euler--Maclaurin formula. – mickep Mar 29 '16 at 19:07
• This should also work for the serie $\sum_{n=1}^{\infty}{\frac{\sin(\sqrt{n})}{n}}$, shouldn't it? – joseph Mar 29 '16 at 19:18
• @joseph yeah sure, this method is quiet flexible and worth knowing especially because it gives us not only the answer to ur question but also an asymptotic value for the sum. – tired Mar 29 '16 at 19:57
• @mickep, yepp u are right, thx for spotting and for the kind words. have a nice evening! – tired Mar 29 '16 at 19:58
• May be interesting to you $$\int_1^N \frac{\sin^2(\sqrt{x})}{x}\,dx=-\text{Ci}\left(2 \sqrt{N}\right)+\text{Ci}(2)+\frac{\log (N)}{2}$$ – Claude Leibovici Jan 15 '17 at 7:37

For $k\in \Bbb N$ let $n_k=\lfloor (k+\frac12)^2\pi^2\rfloor$. Then $\sin^2 \sqrt n_k\approx 1$ and also $\sin^2 \sqrt {n_k+d}\approx 1$ for $0<d<\sqrt{n_d}$. This gives us $\approx k\pi$ summands of size $\approx \frac1{k^2\pi^2}$, i.e., a contribution of $\approx \frac1{k\pi}$. This allows us to compare with the divergent harmonic series.

While the "$\approx$" used in this argument should be made more explicit for a formal proof, we can be quite generous at this ...