# Question #c0aff

Mar 8, 2016

${\sum}_{n = 0}^{\infty} \frac{{2}^{n} {\sin}^{n} \left(x\right)}{n} ^ 2$ converges if and only if

$x \in \left[k \pi - \frac{\pi}{6} , k \pi + \frac{\pi}{6}\right]$ for some $k \in \mathbb{Z}$

#### Explanation:

First, note that $\frac{{2}^{n} {\sin}^{n} \left(x\right)}{n} ^ 2 = {\left(2 \sin \left(x\right)\right)}^{n} / {n}^{2}$.

If $| 2 \sin \left(x\right) | \le 1$ then $| {\left(2 \sin \left(x\right)\right)}^{n} / {n}^{2} | \le \frac{1}{n} ^ 2$.

As ${\sum}_{n = 0}^{\infty} \frac{1}{n} ^ 2$ is a known convergent series, this implies ${\sum}_{n = 0}^{\infty} {\left(2 \sin \left(x\right)\right)}^{n} / {n}^{2}$ converges by the comparison test.

If $| 2 \sin \left(x\right) | > 1$ then ${\lim}_{n \to \infty} | {\left(2 \sin \left(x\right)\right)}^{n} / {n}^{2} |$ diverges (this can be demonstrated through two applications of L'Hopital's rule ) and thus ${\sum}_{n = 0}^{\infty} {\left(2 \sin \left(x\right)\right)}^{n} / {n}^{2}$ diverges by the divergence test.

By the above, we can say that ${\sum}_{n = 0}^{\infty} \frac{{2}^{n} {\sin}^{n} \left(x\right)}{n} ^ 2$ converges if and only if $| 2 \sin \left(x\right) | \le 1$, that is, when $| \sin \left(x\right) | \le \frac{1}{2}$.

Solving for $x$ gives us our final result:

${\sum}_{n = 0}^{\infty} \frac{{2}^{n} {\sin}^{n} \left(x\right)}{n} ^ 2$ converges if and only if

$x \in \left[k \pi - \frac{\pi}{6} , k \pi + \frac{\pi}{6}\right]$ for some $k \in \mathbb{Z}$