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.