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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.

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  • $\begingroup$ do you mean rational by 'e' or the 'e=2.7..' $\endgroup$ – RE60K Aug 16 '14 at 11:10
  • $\begingroup$ Hi,I mean 'e'the base of naural logarithm (2.718281...) $\endgroup$ – M.R. Yegan Aug 16 '14 at 14:40
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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: $$\sin(n!\pi\theta)\to0,\;n\to\infty,\;\theta\in\mathbb Q$$


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 $\bf \theta=e$ .

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  • $\begingroup$ see this for why $\sin(n!\pi e) \to 0$ $\endgroup$ – Petite Etincelle Aug 16 '14 at 11:21
  • $\begingroup$ @LiuGang it is for $n \sin (2 \pi \cdot e \cdot n!)$ $\endgroup$ – RE60K Aug 16 '14 at 11:24
  • $\begingroup$ the technique in that proof applies here too $\endgroup$ – Petite Etincelle Aug 16 '14 at 11:25
  • $\begingroup$ @LiuGang ok my bad, sorry $\endgroup$ – RE60K Aug 16 '14 at 11:38
  • $\begingroup$ Good to write down your thoughts, then we can check for you :) $\endgroup$ – Petite Etincelle Aug 16 '14 at 11:41

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