Divergence-Convergence of the sequence $\sin(n!{\pi}\theta)$

Question

I am working on the convergence-Divergence of $\sin(n!{\pi}\theta).$ In his book, Hardy(A Course of Pure Mathematics) page 128 cited " The case in which $\theta$ is irrational cannot be dealt with without the aid of considerations of a much more difficult character".If $\theta~$is $e$ it can be proved relatively simple that $\sin(n!{\pi}e)~ $approaches $0$ as $n$ tends to $\infty$. Is there any idea or reference about Hardy's say or the general case of $\theta~$. Any hint will be appreciated.

Answer 1

In case of $\theta$ is rational:

Already $$n!=n\times(n-1)\times(n-2)\times\cdots\times1$$ which actually contain product of all natural numbers ($\ge1$) when $n\to\infty$.

So, let $\large\theta=\frac pq$ with $\large p,q\in\mathbb Z,q\ne0$ where p and q are co-prime.

Now q must be an integer, so that it possibly cannot contain product of all the natural numbers.

So, $n!\theta$ will be integer such that $n!\pi\theta$ becomes an integral multiple of $\pi$, consequently: $$\sin(n!\pi\theta)\to0,\;n\to\infty,\;\theta\in\mathbb Q$$

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Instead he said: $$\phi(n)=\sin(n\theta\pi)$$

• The case in which is irrational is a little more difficult. But it is not difficult to see that $\phi(n)$ still oscillates finitely.(After some proof)...Thus the hypothesis that $\phi(n)$ tends to a limit $l$ is impossible, and therefore $\phi(n)$ oscillates as n tends to $\infty$.

• You are considering a different pathway when $n=m!,m\in\mathbb Q$.If there existed a limit for this functoion, it would be independent of path.For your function,there exists a limit for $\bf \theta=e$ .