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

Question

Find values of $x$ for which the following series converges $$ \sum_{n=1}^{\infty} (-1)^n \dfrac{2^n \sin ^{2n}x }{n } $$

Attempt:

(a) Check for Absolute Convergence

If we consider $ \sum_{n=1}^{\infty} (-1)^n \dfrac{2^n \sin ^{2n}x }{n } $, then : $ \sum_{n=1}^{\infty} \sin ^{2n}x $ is bounded and $\le \dfrac {1}{|\sin x|}$.

But $\lim_{n \rightarrow \infty} \dfrac {2^n}{n} = \infty$.

Hence, we can't apply neither the Abel's Test nor the Dirichlets test.

This series does not seem to be absolutely convergent.

(b) Checking for conditional convergence

If $ b_n= (-1)^n \dfrac{2^n \sin ^{2n}x }{n } $, then $\lim_{n \rightarrow \infty} b_n =\infty$.

Hence, we can't even apply he leibnitz condition for the conditional convergence of the series.

Could anyone please help me get a direction.

Thank you very much for your help.

Answer 1

The series converges absolutely if $|\sin x| < \frac{1}{\sqrt{2}}$. Indeed, if $|\sin x| < \frac{1}{\sqrt{2}}$, then $$\lim_{n\to \infty} \left\lvert (-1)^n \frac{2^n\sin^{2n} x}{n}\right\rvert^{1/n} = \lim_{n\to \infty} \frac{2\sin^2 x}{n^{1/n}} = 2\sin^2 x < 1$$ Thus, by the root test, $\sum_{n = 1}^\infty (-1)^n \frac{2^n\sin^{2n} x}{n}$ converges absolutely.

Answer 2

We have $$\sum_{r=1}^\infty\frac{\left(-2\sin^2x\right)^n}n=-\ln[1-(-2\sin^2x)]$$

as $\ln(1-y)=-\sum_{r=1}^\infty\dfrac{y^r}r$

See Taylor series for $\log(1+x)$ and its convergence and Convergence for log 2