TITLE: Is the series $\sum \sin^n(n)$ divergent?
QUESTION [9 upvotes]: I'm almost sure that the series $\sum \sin^n(n)$ is not convergent, but lack proof.
Thank for any help.

REPLY [14 votes]: Using a continued fraction approximation, we know that for any positive integer $N$, we can find integers $q>N$ and $p$ so that
$$
\left|p-q\frac\pi2\right|<\frac1q\tag{1}
$$
It is also true that no two consecutive denominators can share a common factor. Therefore, if one approximation has an even denominator, the next must be odd. So assume that $q$ is odd and $(1)$ is true.  Then, since $\sin\left(q\frac\pi2\right)=(-1)^{(q-1)/2}$ and $\cos\left(q\frac\pi2\right)=0$ the Maclaurin Series yields
$$
\begin{align}
(-1)^{(q-1)/2}\sin(p)
&\ge1-\frac12\left(p-q\frac\pi2\right)^2\\
&\ge1-\frac{1}{2q^2}\tag{2}
\end{align}
$$
Thus, for continued fraction approximations $\frac{p}{q}$ to $\frac\pi2$ with odd denominators $|\sin(p)|\ge1-\frac{1}{2q^2}$. Taking the $\liminf$ as $q\to\infty$ yields
$$
\begin{align}
\liminf_{p\to\infty}|\sin(p)|^p
&\ge\lim_{p\to\infty}\left(1-\frac{1}{2q^2}\right)^p\\
&=\lim_{q\to\infty}\left(1-\frac{1}{2q^2}\right)^{q\pi/2}\\
&=1\tag{3}
\end{align}
$$
Inequality $(3)$ implies that
$$
\limsup_{n\to\infty}|\sin^n(n)|=1\tag{4}
$$
Since the terms do not tend to $0$,
$$
\sum_{n=0}^\infty\sin^n(n)
$$
does not converge.