Convergence of the series $\sum\limits_{n=1}^{\infty }(-1)^{n+1} \frac{\sin^2(n)}{n}$ I would like to know how to prove the convergence (or not) of the following serie:
$\sum\limits_{n=1}^{\infty }(-1)^{n+1} \frac{\sin^2(n)}{n}$
Thank you in advance for any suggestion.
 A: $\sin^2(n) = {1 - \cos(2n) \over 2}$, so your series is $\sum_n (-1)^{n+1}({1 - \cos(2n) \over 2n})$. Since
the sum of ${(-1)^{n+1} \over n}$ converges to $\ln(2)$, it suffices to show that 
$\sum_n {(-1)^n \cos(2n) \over 2n}$ converges. This is the real part of $\sum_n {(-1)^n e^{2in} \over 2n}$.
The series $\ln(1 + z) = -\sum_n {(-1)^n z^n \over n}$ converges for all $|z| \leq 1$ other than $z = 1$
(This is pretty standard and can be shown using Dirichlet's test). Plugging in
$z = e^{2i}$ this gives that the series converges. Thus so does your original series. 
A: You can not use the Leibniz Criterion for this series because the term $\frac{\sin^2{n}}{n}$ is not monotone decreasing (Look at n = 3 and n = 4).
Intuitively, we know that $0 \leq \sin^2{n} \leq 1$, and we know something about the convergence behavior of $\sum{\frac{(-1)^{n+1}}{n}}$, so it isn't completely unreasonable to expect this series to converge. Now, to actually prove its convergence...
HINT: Think about how you can apply Dirichlet's Test.
A: $\sum_{n=1}^{N}(-1)^{n+1}\sin^2(n)=\sum_{n=1}^{N}(-1)^{n+1}\frac{1 - \cos(2n)}{2}=\frac{1}{2}\Re(\sum_{n=1}^{N}(-1)^n e^{i2n})+\sum_{n=1}^{N}\frac{(-1)^{n+1}}{2}$
$$\left|\frac{1}{2}\Re(\sum_{n=1}^{N}(-1)^n e^{i2n})+\sum_{n=1}^{N}\frac{(-1)^{n+1}}{2}\right|\leq\left|\frac{1}{2}\Re(\sum_{n=1}^{N}(-1)^n e^{i2n})\right|+\left|\sum_{n=1}^{N}\frac{(-1)^{n+1}}{2}\right|$$
$$\left|\sum_{n=1}^{N}\frac{(-1)^{n+1}}{2}\right|\leq \frac{1}{2}\ \ \ , \ \ \ \left|\frac{1}{2}\Re(\sum_{n=1}^{N}(-1)^n e^{i2n})\right|\leq\left|\sum_{n=1}^{N}(-1)^n e^{i2n}\right|<\frac{2}{1}$$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sum_{n=1}^{N}(-1)^n e^{i2n}=a_1\cdot\frac{1-r^N}{1-r},\ r=-e^{2i}=-\cos(2)-i\sin(2),\ a_1=r\ \ \ \ \ \ \ \ \ \ \ \ \ $(*)
Which means that the sum of the series $\sum_{n=1}^{N}(-1)^{n+1}\sin^2(n)$ is bounded for any $N\in\mathbb{Z}_+$, so if $b_n=(-1)^{n+1}\sin^2(n)$ and $a_n=\frac{1}{n}$, then Dirichlet's Test is passed and that's why the sum of the series $\sum\limits_{n=1}^{\infty }(-1)^{n+1} \frac{\sin^2(n)}{n}$ converges. The exact value of it is here.
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,{\sin^{2}\pars{n} \over n}:\
     {\large ?}}$

With $\ds{\mu\ \in\ \left[0,1\right)}$, let's
  $\ds{{\cal I}\pars{\mu}
     =\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,\mu^{n}\,{\sin^{2}\pars{n} \over n}}$ such that $\ds{\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,{\sin^{2}\pars{n} \over n} = {\cal I}\pars{1^{-}}}$ and $\ds{{\cal I}\pars{0} = 0}$:

\begin{align}
{\cal I}'\pars{\mu}&=\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\mu^{n - 1}\,
{1 - \cos\pars{2n} \over 2}
=\half\sum_{n = 1}^{\infty}\pars{-\mu}^{n - 1}
+ {1 \over 2\mu}\Re\sum_{n = 1}^{\infty}\pars{-\mu\expo{2\ic}}^{n}
\\[3mm]&=\half\,{1 \over 1 + \mu} + {1 \over 2\mu}\,\Re\bracks{
{-\mu\expo{2\ic} \over 1 + \mu\expo{2\ic}}}
=\half\,{1 \over 1 + \mu} - \Re\bracks{{\expo{2\ic} \over 2}\,
{1 \over 1 + \mu\expo{2\ic}}}\,,\qquad{\cal I}\pars{0} = 0
\end{align}

\begin{align}
&\color{#00f}{\large%
\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,{\sin^{2}\pars{n} \over n}}=
\int_{0}^{1^{-}}\braces{\half\,{1 \over 1 + \mu} -
\Re\bracks{{\expo{2\ic} \over 2}\,
{1 \over 1 + \mu\expo{2\ic}}}}\dd\mu
\\[3mm]&=\half\bracks{%
\ln\pars{1 + \mu} - \Re\ln\pars{1 + \mu\expo{2\ic}}}_{0}^{1^{-}}
=\half\bracks{\ln\pars{2} - \Re\ln\pars{1 + \cos\pars{2} + \ic\sin\pars{2}}}
\\[3mm]&=\half\bracks{\ln\pars{2}- \ln\pars{\root{2 + 2\cos\pars{2}}}}
=\half\bracks{\ln\pars{2}- \ln\pars{\root{4\cos^{2}\pars{1}}}}
\\[3mm]&=\half\bracks{\ln\pars{2}- \ln\pars{2\cos\pars{1}}}
=\color{#00f}{\large -\,\half\,\ln\pars{\cos\pars{1}}}
\approx 0.3078
\end{align}

It converges !!!.
