# Showing convergence of $\sum_{n=1}^{\infty} {(-1)^{n} \frac{\sin^{2} n}{n}}$

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?

• 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? – Conrad Feb 10 at 15:58

$$\sum_{n=1}^{m} (-1)^{n} \frac{\sin^{2} n}{n} = \sum_{n=1}^{m} \frac{ (-1)^{n}}{2n} - \sum_{n=1}^{m}\frac{ (-1)^{n}\cos 2n}{2n} \\= \sum_{n=1}^{m} \frac{ (-1)^{n}}{2n} - \sum_{n=1}^{m}\frac{ \,\cos (\pi+2)n}{2n}$$
The partial sums on the RHS are both convergent as $$m \to \infty$$.
• $\left|\sum_{n=1}^m \cos \alpha n \right| \leqslant \frac{1}{\sin(\alpha/2)}$ if $\alpha$ is not an integer multiple of $2\pi$. – RRL Feb 10 at 16:16