For any fixed $m\ge2$, find the value of $$\lim_{n\to\infty}\sum_{k=1}^{m}\sin^{n}\left(\frac{k!\pi}{m}\right)+\cos^{n}\left(\frac{k!\pi}{m}\right).$$
Is there a formula for $\sin^{n}\theta+\cos^{n}\theta$ for any $n\in\mathbb{N}$?.
I can't find any method to solve above problem.
Give some advice! Thank you!
This is a partial answer, not because the full answer is unsolved, but because it’s unclear what type of answer you’re looking for. This feels to me more like a programming problem than a math problem.
If $m$ is odd, then $k!/m$ is never a half-integer. Therefore the $\sin$ terms always converge to $0$. Furthermore since $m>1$, $k!$ will be even for any $k$ such that $k!/m$ is an integer, so the $\cos$ terms are either identically $1$ or converge to $0$. We simply need to count how many $1$-terms in the sum: that is, when is $k!/m$ an even integer?
It is clear that there is some least integer $k_0$ for which $k_0!/m$ is an (even) integer, and all subsequent values are also even integers. Hence the value of the sum is just $m - k_0 + 1$. The precise value of $k_0$ turns out to be closely related to the factorization of $m$ into prime powers: each prime power $p^r$ which divides $m$ yields a lower bound for $k_0$ (roughly equal to $r(p-1)$ but whose exact value can be obtained from Lucas’s theorem). Then $k_0$ is just the max of all such lower bounds.
The case of $m$ even is a little more complicated but the analysis is substantially similar. There will be some cases where there is at least one value of $\sin$ or $\cos$ equal to $-1$, and for these the limit does not exist. In other cases it still boils down to counting $1$s, but here there may be a couple $\sin$ terms that give $1$ in addition to the $\cos$.