Convergence of the series $\sum\limits_{n=1}^{\infty }(-1)^{n+1} \frac{\sin^2(n)}{n}$

Question

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.

Answer 1

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

Answer 2

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.