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Find all x such that the series (-1)n (2n sin2nx) / n converges

by RoRi

Find all real numbers x such that the series

    \[ \sum_{n=1}^{\infty} (-1)^n \frac{2^n \sin^{2n} x }{n}\]

converges.

We have

    \[ \sum_{n=1}^{\infty} (-1)^n \frac{2^n \sin^{2n} x}{n} = \sum_{n=1}^{\infty} \frac{1}{n}(-2 \sin^2 x)^n. \]

This series converges if and only if 2 \sin^2 x \leq 1.

    \begin{align*} 2 \sin^2 x \leq 1 && \implies && \sin^2 x &\leq \frac{1}{2} \\[9pt] && \implies && \sin x &\leq \frac{\sqrt{2}}{2} \\[9pt] && \implies && |x-k\pi| &\leq \frac{\pi}{4} \end{align*}

for k \in \mathbb{Z}.

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