# Convergence of the following sum

Does the following sum converge? $$\sum_{n=1}^{\infty}\frac{\sin^2(n)}{n}$$ I tried the ratio test and got that $\rho=0$ which means that the series converges absolutely. However, Mathematica and Wolfram Alpha do not give a result when trying to find its convergence. Am I wrong?

• Your ratio test result is not correct. Can you show your work for it? Mar 2, 2017 at 2:25
• I eveluated the following limit: $\lim_{n\to\infty}\left|\frac{n\sin^2(n+1)}{(n+1)\sin^2(n)}\right|$ which by wolfram alpha is also 0
– user372003
Mar 2, 2017 at 2:31
• You must have typed in the wrong expression. The limit you specified above does not exist. An answer of zero is silly -- what could force it to be zero? Mar 2, 2017 at 2:35
• Actually it does seem to be a bug in Wolfram Alpha. Another reminder to think before you compute. Mar 2, 2017 at 2:37
• @RobertIsrael What should be the output of the limit then? As I can see from its graph it does not seem to exist
– user372003
Mar 2, 2017 at 5:55

Yes, you are wrong. The ratio test is inconclusive, and the series diverges.

Note that there is some $\varepsilon > 0$ such that $\sin^2(n) + \sin^2(n+1) > \varepsilon$ for all $n$. This is because if $n$ is close to a multiple of $\pi$, $n+1$ will not be. Thus $$\frac{\sin^2(n)}{n} + \frac{\sin^2(n+1)}{n+1} \ge \frac{\varepsilon}{n+1}$$ and a comparison with the harmonic series shows that the series diverges.

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 1}^{N}{\sin^{2}\pars{n} \over n} & = {1 \over 2}\sum_{n = 1}^{N}{1 \over n} - {1 \over 2}\,\Re\sum_{n = 1}^{N}{\exp\pars{2\ic n} \over n} \end{align}

But,

$\ds{{1 \over 2}\,\Re\sum_{n = 1}^{\infty}{\exp\pars{2\ic n} \over n} = -\,{1 \over 2}\,\Re\ln\pars{1 - \exp\pars{2\ic}} = -\,{1 \over 2} \ln\pars{\root{\bracks{1 - \cos\pars{2}}^{\,2} + \sin^{2}\pars{2}}} = -\,{1 \over 4}\ln\pars{2\bracks{1 - \cos\pars{2}}} = -\,{1 \over 4}\ln\pars{4\sin^{2}\pars{1}} = \bbx{\ds{-\,{1 \over 2}\,\ln\pars{2\sin\pars{1}}}}}$

So, $$\bbx{\ds{% \sum_{n = 1}^{N}{\sin^{2}\pars{n} \over n} \sim {1 \over 2}\,H_{N} + {1 \over 2}\,\ln\pars{2\sin\pars{1}} \qquad\mbox{as}\ N \to \infty}}$$

WA is obviously incorrect. To show this, recall that $\sin^2(n)=\frac12-\frac12\cos(2n)$.

Then, letting $S_N=\sum_{n=1}^N \frac{\sin^2(n)}{n}$, $H_N=\sum_{n=1}^N\frac1n$, and $T_N=\sum_{n=1}^N\frac{\cos(2n)}{n}$, it is easy to see that

\begin{align} S_N+\frac12 T_N &=\frac12 H_N\end{align}

Note that Dirichlet's Test guarantees that $T_N$ converges. Hence, if we assume that $S_N$ converges, then the sum $S_N+\frac12 T_N=H_N$ converges.

Inasmuch as the Harmonic series diverges, this leads to a contradiction and we find that $S_N$ diverges. And we are done.

• Please let me know how I can improve this answer too. As always, I really want to give you the best answer I can. If the answer was not useful, I am happy to delete it. -Mark Apr 12, 2017 at 17:16