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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.

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Belongs on math.stackexchange.com –  Platinum Azure Mar 17 '11 at 15:18
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migrated from stackoverflow.com Mar 19 '11 at 2:50

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4 Answers

$\sin^2(n) = {1 - \cos(2n) \over 2}$, 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.

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I am not sure if you get full number of points if you solve the problem like this during a test. –  Darius Feb 23 at 10:58
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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.

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I am not exactly sure if Dirichlet test will be of help as it is... –  user17762 Mar 19 '11 at 6:11
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$\sum_{n=1}^{N}(-1)^{n+1}\sin^2(n)=\sum_{n=1}^{N}(-1)^{n+1}\frac{1 - \cos(2n)}{2}=\frac{1}{2}\Re(\sum_{n=1}^{N}(-1)^n e^{i2n})+\sum_{n=1}^{N}\frac{(-1)^{n+1}}{2}$ $$\left|\frac{1}{2}\Re(\sum_{n=1}^{N}(-1)^n e^{i2n})+\sum_{n=1}^{N}\frac{(-1)^{n+1}}{2}\right|\leq\left|\frac{1}{2}\Re(\sum_{n=1}^{N}(-1)^n e^{i2n})\right|+\left|\sum_{n=1}^{N}\frac{(-1)^{n+1}}{2}\right|$$ $$\left|\sum_{n=1}^{N}\frac{(-1)^{n+1}}{2}\right|\leq \frac{1}{2}\ \ \ , \ \ \ \left|\frac{1}{2}\Re(\sum_{n=1}^{N}(-1)^n e^{i2n})\right|\leq\left|\sum_{n=1}^{N}(-1)^n e^{i2n}\right|<\frac{2}{1}$$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \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.

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,{\sin^{2}\pars{n} \over n}:\ {\large ?}}$

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 !!!.

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And what about convergence of $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\sin^2(n)}{\sqrt{n}}$? –  Darius Feb 23 at 11:02
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