# 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.

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Belongs on math.stackexchange.com –  Platinum Azure Mar 17 '11 at 15:18

## migrated from stackoverflow.comMar 19 '11 at 2:50

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