Convergence of the series $\sum_{n=0}^\infty \sin(n! \pi m \sin(1))$

Question

In this exercise I was asked to prove the convergence of the following infinite sum: $$\sum_{n=0}^\infty \sin(n! \pi m \sin(1)),$$ where $m$ denotes any integer. I don't have any idea on how to approach such problem, so I would be very grateful if you could explain your solution.

Answer 1

If $n$ is odd, then

$$\pi n!\sin 1 = \pi n!(1 - 1/3! + 1/5! - \cdots ) = \pi\cdot (\text {integer} + r_n),$$

where

$$|r_n|< 1/[(n+2)(n+1)] + 1/[(n+4)(n+3)(n+2)(n+1)] + \cdots$$ $$ < 1/[(n+2)(n+1)] + 1/[(n+2)(n+1)]^2 + 1/[(n+2)(n+1)]^3 + \cdots = 1/[(n+2)(n+1)-1].$$

It follows that for large $n$ odd,

$$ |\sin (\pi n!\sin 1)| < \pi/[(n+2)(n+1)-1]$$

for large $n.$ So if we were summing over just odd $n,$ we would have absolute convergence. But we have the even terms to worry about, and those seem more subtle. So there's more work to do, but this is a start.