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So I need to evaluate $$\lim_{n\to\infty}n\sin(2\pi en!)$$ And yes, I know it was discussed here and similar limit was discussed here, but I didn't feel quite okay with the solutions/hints given in those links so I did it my way. That's what I did: $$\begin{align}\lim_{n\to\infty}n\sin(2\pi en!)&=^{stirling}\lim_{n\to\infty}n\sin\left(2\pi e \sqrt{2\pi n} \left(\frac{n}{e}\right)^n\right)\\ &=\lim_{n\to\infty}n\sin\left((2\pi)^{\frac{3}{2}}e^{1-n}n^{n+\frac{1}{2}}\right)\\ &=^{\{en!\}=\frac{1+O(n^-1)}{n+1}}\lim_{n\to\infty}n\sin\left(\frac{2\pi (1+O(n^{-1})}{n+1}\right)\\ &=^{sin(n^{-1})\sim n^{-1}}\lim_{n\to\infty}\sin\left(\frac{2\pi n (1+O(n^{-1})}{n+1}\right)\\ &=2\pi\end{align}$$ But my professor didn't accept it because we didn't cover Stirling yet and said I need to prove it another way (he also didn't accept the solutions given in the links above).
He told me to use the fact that $$\lim_{n\to\infty}n\sin(2\pi en!)=\lim_{n\to\infty}n\sin(2\pi en!-2\pi n)$$ and then using Taylor or something, but this so called "clue" got me nowhere.
Any ideas?

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  • Doesn't the n dominate the sin which stays bounded between 1 and -1? so the limit if it exists would be either inf or -inf? – Mohamed Alaa El Behairy Feb 01 '15 at 19:32
  • No, see the links :) – user114138 Feb 01 '15 at 19:52
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    Your use of Stirling's approximation here is both unnecessary (you don't actually make use of it) and incorrect (the two limits are not equal). The error term in Stirling is much, much too large to detect the value of $\sin(2\pi e n!)$. – Erick Wong Feb 01 '15 at 19:52
  • Are you sure? My proffesor said it's right, plus I've seen similar proof whilst searching alternative solution – user114138 Feb 01 '15 at 20:02
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    @user114138 Yes, very sure. The proof is only correct if you delete the first two equalities and omit any reference to Stirling's. It's as if you wrote $x^2-y^2 = \mbox{zebra} = \mbox{four-legged creature} = (x-y)(x+y)$. The end result is correct but the part in between is irrelevant nonsense. – Erick Wong Feb 04 '15 at 12:53
  • @user114138 As illustration, try $n=10$. The value of $10!$ is $3628800$. The value of Stirling's is $3598695.618$, a very nice approximation with an error of over $30000$. In order to even come remotely close to estimating $\sin (2\pi e n!)$ you would need that deviation to be less than $1/2e$. It's a hundred thousand times off that mark and it only gets further away as $n$ increases. Try it yourself. – Erick Wong Feb 04 '15 at 13:11
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    Stirling only works if you know the additive error. That first step is simply unsupported by basic Stirling formula. – Thomas Andrews Apr 12 '15 at 17:46

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