An interesting limit

**zadir** · November 22nd 2010, 10:44 AM

$\lim \{n!e\}= \:?$
where {.} is the symbol of the fractinal part

Thank you in advance!

**roninpro** · November 22nd 2010, 11:00 AM

I think that the important observation is that $\displaystyle e=\sum_{k=1}^\infty \frac{1}{k!}$ , so $\displaystyle n!\sum_{k=1}^\infty \frac{1}{k!}=\sum_{k=1}^\infty \frac{n!}{k!}$ will be very close an integer when $n$ is large. So, $\{n!e\}\to 0$ .

I haven't thought it all the way through, but I think that you can make it rigorous by looking at the partial sums of the series and using Taylor's theorem. Give it a try.

**tonio** · November 22nd 2010, 11:04 AM

Originally Posted by zadir

$\lim \{n!e\}= \:?$
where {.} is the symbol of the fractinal part

Thank you in advance!

$\displaystyle{e=\sum\limits^\infty_{k=0}\frac{1}{k !}\Longrightarrow n!e=n!\left(1+1+\frac{1}{2!}+\ldots+\frac{1}{n!}+\ sum\limits^\infty_{k=n+1}\frac{1}{k!}\right)}$ $\Longrightarrow\displaystyle{\{n!e\}=\sum\limits_{ k=n+1}^\infty\frac{n!}{k!}}$

Tonio

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