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What will be the value of this limit?

$$\lim_{n\to\infty}n\cdot\sin(2\pi\ e\ n!)$$

Any help on how to proceed?

Ali Caglayan
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2 Answers2

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Consider that, as $n$ increases, the distance of $e n!$ from the closest integer drops to zero, since: $$ n!e=n!\cdot\sum_{k=0}^{+\infty}\frac{1}{k!}=A+\frac{1}{(n+1)}+\frac{1}{(n+1)(n+2)}+\ldots $$ where $A\in\mathbb{N}$. This gives: $$ \{n! e\} = \frac{1}{n+1}+O\left(\frac{1}{n^2}\right)$$ and: $$n\sin(2\pi n! e) = n\sin\left(2\pi\left(\frac{1}{n+1}+O\left(\frac{1}{n^2}\right)\right)\right),$$ so the limit is $2\pi$.

Jack D'Aurizio
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  • Did you expand $\sin (a+b)$ in the last step? – Alex Dec 06 '14 at 16:34
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    I just used $\sin(2\pi(A+u))=\sin(2\pi A+2\pi u)=\sin(2\pi u)$ that comes from $A\in\mathbb{N}$. – Jack D'Aurizio Dec 06 '14 at 16:54
  • I'm sorry, what's A here? $\frac{1}{n+1}$? – Alex Dec 06 '14 at 19:02
  • @Alex: $$A = n!\sum_{k=0}^{n}\frac{1}{k!}\in\mathbb{N}.$$ – Jack D'Aurizio Dec 06 '14 at 19:08
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    @JackDAurizio Correct me if I'm wrong, but I think since $n$ tends to infinity, $\sin(2\pi e n!)$ will oscillate between $1$ and $-1$ whereas $n$ will tend to infinity so that the limit also tends to infinity. Thanks! – Henry Dec 06 '14 at 19:59
  • @JackD'Aurizio: I'm sorry, my question is how you got the limit from $\sin(2 \pi (\frac{1}{n+1} + O(\cdot))$? Did you use the $\sin (a + b) = \sin a \cos b + \cos a \sin b$ expansion? – Alex Dec 06 '14 at 20:10
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    For both of you: we have $\sin(2\pi(A+u))=\sin(2\pi u)$, where $u$ is very small. Since $$\lim_{z\to 0}\frac{\sin z}{z}=1,$$ the claim follows. – Jack D'Aurizio Dec 07 '14 at 00:10
4

Expand $e=1+\frac11+\frac1{2!}+...+\frac1{n!}+...$

Empy2
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