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I've tried to prove the convergence of $$\sum_{n=1}^{\infty} {(-1)^{n} \frac{\sin^{2} n}{n}}$$ using three different criteria but ran into the following issues:

$1.$ I first tried the Leibnitz criterion.

Let $a_n = \displaystyle{\frac{\sin^{2} n}{n}}$. Though $$\lim_{n\to \infty} \frac{\sin^{2} n}{n} = 0$$ the sequence $a_n$ is not monotone thus the test fails.

$2.$ Next I tried Dirichlet's test. I set $\displaystyle{ {a_n} = \frac{1}{n}}$ which monotonically decreases and has a limit of zero. But with $\displaystyle{{b_n} = (-1)^n \sin^2{n}}$ the partial sums $\displaystyle{\left|\sum_{n=1}^{m}b_n\right|}$ are not bounded. Same issue with setting $\displaystyle{ a_n = \frac{(-1)^n}{n}}$ and $\displaystyle{ b_n = \sin^2 n }$.

$3.$ The last criterion I know is Abel's test. I'm stuck on similar points as with Dirichlet's test. If I set $\displaystyle{ a_n = \frac{\sin^{2} n}{n} }$, then $\displaystyle{ \sum {a_n} }$ is convergent, but what remains $\displaystyle{ (-1)^n }$ is bounded but not monotone.

If $ \displaystyle{ a_n = \frac{(-1)^n}{n} }$ then $\displaystyle{ \sum {a_n} }$ is convergent, but $\displaystyle{ \sin^2 n }$ is not monotone.

What other tests can I apply? Or perhaps this shows that the series is not convergent?

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    $\begingroup$ what about if you use the double angle formula and express the sin square in terms of 1 and cos and see if you can apply the usual stuff as above? $\endgroup$
    – Conrad
    Commented Feb 10, 2019 at 15:58

1 Answer 1

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Hint:

$$\sum_{n=1}^{m} (-1)^{n} \frac{\sin^{2} n}{n} = \sum_{n=1}^{m} \frac{ (-1)^{n}}{2n} - \sum_{n=1}^{m}\frac{ (-1)^{n}\cos 2n}{2n} \\= \sum_{n=1}^{m} \frac{ (-1)^{n}}{2n} - \sum_{n=1}^{m}\frac{ \,\cos (\pi+2)n}{2n} $$

The partial sums on the RHS are both convergent as $m \to \infty$.

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    $\begingroup$ $\left|\sum_{n=1}^m \cos \alpha n \right| \leqslant \frac{1}{\sin(\alpha/2)}$ if $\alpha$ is not an integer multiple of $2\pi$. $\endgroup$
    – RRL
    Commented Feb 10, 2019 at 16:16

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