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show that

$$\lim_{n\to\infty}n^2\left(n\sin{(2e\pi\cdot n!)}-2\pi\right)=\dfrac{2\pi(2\pi^2-3)}{3}$$

we are kown that $$\lim_{n\to\infty}n\sin{(2\pi e\cdot n!)}=2\pi$$

because we note $$e=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots+\dfrac{1}{n!}+\dfrac{1}{(n+1)!}+O(\dfrac{1}{(n+1)!})$$

then $$2e\pi\cdot n! =2k\pi+\dfrac{2\pi}{n+1}+o(\dfrac{1}{n+1}))$$

for this problem we $$e=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots+\dfrac{1}{n!}+\dfrac{1}{(n+1)!}+\dfrac{1}{(n+2)!}+o(\dfrac{1}{(n+2)!})$$ so $$2\pi en!=2k\pi+\dfrac{2\pi}{n+1}+\dfrac{2\pi}{(n+1)(n+2)}+o(\dfrac{1}{(n+1)(n+2)})$$

and use $$\sin{x}=x-\dfrac{x^3}{6}+o(x^3)$$ But I can't work

math110
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  • Seems like you know the right idea if you know how to get what you say you know. Consider the next term in the Taylor series of $\sin$? – Ted Shifrin Sep 13 '13 at 02:04
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    Sorry to question what you've written, but could you double check everything? It doesn't seem your limit is even close to $\frac{2}{3}\pi(2\pi^2-3)$ – L. F. Sep 13 '13 at 02:09

1 Answers1

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Clearly the fractional part of $\mathrm{e} \cdot n!$ equals to $$\begin{eqnarray} \{\mathrm{e} \cdot n!\} &=& \left\{\sum_{k=0}^{n} \frac{n!}{k!} + \sum_{k=0}^\infty \frac{n!}{(n+1+k)!}\right\} =\left\{ \sum_{k=0}^\infty \frac{n!}{(n+1+k)!}\right\} = \left\{ \sum_{k=0}^\infty \frac{1}{(n+1)_{k+1}}\right\} \\ &=& \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \mathcal{o}\left(n^{-3}\right) \end{eqnarray} $$ Hence $$ n \cdot \sin\left(2 \pi \mathcal{e} n!\right) = 2 \pi - \frac{2 \pi}{n^2} \left(1 + \frac{2 \pi^2}{3}\right) + \mathcal{o}\left(n^{-3}\right) $$ Hence $$ \lim_{n \to \infty} n^2 \left( \sin\left(2 \pi \cdot \mathcal{e}\cdot n!\right) - 2 \pi \right) = - 2 \pi \left(1 + \frac{2 \pi^2}{3}\right) \approx -47.6248875475793{\color \gray{467109}}$$ Here is a numerical confirmation: enter image description here

Sasha
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