Find all real numbers 
such that the series
![\[ \sum_{n=1}^{\infty} (-1)^n \frac{2^n \sin^n x }{n^2}\]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-432e69282fb3807fa8bd3c92bd446d3c_l3.png "Rendered by QuickLaTeX.com")
converges.
We have
![\[ \sum_{n=1}^{\infty} \frac{2^n \sin^n x}{n^2} = \sum_{n=1}^{\infty} \left( \frac{1}{n^2} \right) \left( 2 \sin x \right)^n. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a2f83bcee0b1ddc02903b06905791352_l3.png "Rendered by QuickLaTeX.com")
We know 
conveges; therefore, the product converges if and only if the geometric series 
converges. For this we must have
![\[ 2 \sin x \leq 1 \quad \implies \quad \sin x \leq \frac{1}{2} \quad \implies \quad |x-k \pi|\leq \frac{\pi}{6} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ba4f2224b66c55c9051325c819e94b68_l3.png "Rendered by QuickLaTeX.com")
for 
.
Find all real numbers such that the series
converges.
We have
We know conveges; therefore, the product converges if and only if the geometric series converges. For this we must have
for .