Does $\sum_{n=1}^{\infty}(\sin{n})^n$ converge?

Question

As above, does $\sum_{n=1}^{\infty}(\sin{n})^n$ converge? And if so, to what value.

From calculating partial sums, it appears it might, but I'm not quite sure how to proceed from there.

Answer 1

In Khinchin's little book on continued fractions the following theorem (Theorem 24) due to Chebyshev is given: For an irrational $\alpha$ and any $\beta$ there are infinitely many pairs of integers $x>0,y$ such that $|\alpha x-y-\beta|<\frac{3}{x}$.

We apply this for $\alpha=2\pi,\beta=-\frac{\pi}{2}$, and for any solution $x,y$ we get $$|1-\sin y|=|\sin(\alpha x-\beta)-\sin y|\leq|\alpha x-\beta-y|<\frac{3}{y}$$ (we use the inequality $|\sin a-\sin b|\leq|a-b|$, which follows from $|\sin'x|\leq 1$ and the mean value theorem), so $\sin y\geq 1-\frac{3}{y}$, $(\sin y)^y\geq\left(1-\frac{3}{y}\right)^y\to e^{-3}>0$ as $y\to\infty$. Therefore $(\sin n)^n$ doesn't converge to zero and the series doesn't converge.