$\sum_{n=1}^\infty\frac{(-1)^n\sin ^2n}{n}$ Is the following solution wrong ?; Does $\sum\frac{(-1)^n\cos 2n}{2n}$ converge?

Question

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

Solution from the lecture notes :

$$\frac{(-1)^n\sin ^2n}{n}=\frac{(-1)^n(1-\cos > 2n)}{2n}=\frac{(-1)^n}{2n}-\frac{(-1)^n\cos 2n}{2n}$$ $\frac{(-1)^n}{2n}$ converges conditionally due to Leibniz test.

$\frac{(-1)^n\cos 2n}{2n}$ converges absolutely since $\frac{\cos 2n}{2n}$ converges due to Dirichlet test.

We conclude that the sum we started with coverges conditionally.

$\frac{(-1)^n\cos 2n}{2n}$ cannot converge absolutely because $$\left |\frac{(-1)^n\cos 2n}{2n} \right |=\left |\frac{\cos 2n}{2n} \right |>\frac{\cos^2 2n}{2n}=\frac{\frac12(1+\cos4n)}{2n}=\frac1{4n}+\frac{\cos4n}{4n}$$

This is a sum of a convergent and divergent series hence the series doesn't absolutely converge.

Am I correct and does $\sum\frac{(-1)^n\cos 2n}{2n}$ converge ?

Answer 1

$$\sum_{n\geq 1}\frac{(-1)^n}{n}\cos(2n) $$ does not converge absolutely but it is conditionally convergent by Dirichlet's test. By cheating a bit in evaluating a power series on the boundary of its region of convergence we have

$$\sum_{n\geq 1}\frac{(-1)^n}{n}\cos(2n)=\text{Re}\sum_{n\geq 1}\frac{(-e^{2i})^n}{n}=-\text{Re}\log(1+e^{2i})=\color{red}{\log\left(\frac{\sin 1}{\sin 2}\right)} $$ and this can be turned into an actual proof by invoking summation by parts / Abel's lemma.
It leads to $$ \sum_{n\geq 1}\frac{(-1)^n}{n}\sin^2(n) = \color{red}{\frac{1}{2}\log\cos(1)}$$ and this identity can also be recovered from the Fourier series of $\log\cos$.

Answer 2

Dirichlet test with $b_k = \frac{1}{k}$ monotone decreasing and converging to $0$ and $a_k = (-1)^k \sin^2 k$:

$$ S_M = |\sum_{k=1}^{M}(-1)^k \sin^2(k) |\leq |\sum_{k=1}^{M}(-1)^k | \leq 1\\ $$