Is the series $\sum \sin^n(n)$ divergent? I'm almost sure that the series $\sum \sin^n(n)$ is not convergent, but lack proof.
Thank for any help.
 A: 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.
A: For any irrational $r$ (in particular $\pi/2$) there are infinitely many fractions $p/q$ such that $|r - p/q| < 1/q^2$.  IIRC we can specify that $q$ is odd, perhaps at the cost of changing $1/q^2$ to $c/q^2$ for some constant $c$.  Taking $r = \pi/2$, this says
$|q \pi/2 - p| < c/q$, and then $|\sin p| > 1 - c^2/(2 q^2)$ and $|\sin p|^p > (1-c^2/(2 q^2))^p$.
Now as $q \to \infty$ with $p \approx q \pi/2$, $(1 - c^2/q^2)^p \to 1$.  Thus for any $\epsilon > 0$ there are infinitely many $n$ with $|\sin n|^n > 1 - \epsilon$.
A: Hint: If $\frac{n}{k}$ is a "good" approximation to $\frac{\pi}2$ and $k$ is odd and large, then $|\sin^n n|$ is close to $1$.
