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? 
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}}}
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\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,
$$\bbx{\ds{%
\sum_{n = 1}^{N}{\sin^{2}\pars{n} \over n} \sim
{1 \over 2}\,H_{N} + {1 \over 2}\,\ln\pars{2\sin\pars{1}}
\qquad\mbox{as}\ N \to \infty}}
$$
A: 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. 
