Convergence of: $\sum _{n=1}^{\infty }{\left(-1\right)^{n}\frac{2^{n}\sin^{2n}{\left(x\right)}}{n}}$

Question

I solved that it converges absolutely on $\bigcup \left(-\frac{\pi}{4}+k\pi,\frac{\pi}{4}+k\pi\right) \text{for all k belongs to integer numbers}$ I don't know how to deal with "non-absolute" convergence.I think I can't use alternating series test becuase of non-decreasing character of sequence.Any hints?

$$\sum _{n=1}^{\infty }{\left(-1\right)^{n}\frac{2^{n}\sin^{2n}{\left(x\right)}}{n}}$$

Answer 1

Hint For fixed $x$, the series is $$\sum_{n = 1}^{\infty} (-1)^n \frac{r^n}{n} ,$$ where $r := 2 \sin^2 x$. Show that this converges absolutely for $r < 1$, converges conditionally for $r = 1$, and diverges for $r > 1$.

Additional hints Respectively: Compare with a geometric series with ratio $r$; use the Alternating Series Test; show that the summand diverges as $n \to \infty$.

Answer 2

Consider the power series $\sum_{n=1}^\infty\frac{(-2)^n}ny^n$. By the formula of Cauchy-Hadamard, the radius of convergence is $r=1/2$. So, your series converges absolutely whenever $\sin^2(x) < 1/2$, and thats exactly for $x$ in the set you provide in your question. We also know from the theory of power series that the series does not converge for $|y|>1/2$, hence it remains to consider the case where $y = \sin^2(x)= 1/2$. In this case, the series converges by the Leibniz criterion.