$\displaystyle \lim \{n!e\}= \:?$

where {.} is the symbol of the fractinal part

Thank you in advance!

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- Nov 22nd 2010, 10:44 AMzadirAn interesting limit
$\displaystyle \lim \{n!e\}= \:?$

where {.} is the symbol of the fractinal part

Thank you in advance! - Nov 22nd 2010, 11:00 AMroninpro
I think that the important observation is that $\displaystyle \displaystyle e=\sum_{k=1}^\infty \frac{1}{k!}$, so $\displaystyle \displaystyle n!\sum_{k=1}^\infty \frac{1}{k!}=\sum_{k=1}^\infty \frac{n!}{k!}$ will be very close an integer when $\displaystyle n$ is large. So, $\displaystyle \{n!e\}\to 0$.

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. - Nov 22nd 2010, 11:04 AMtonio
$\displaystyle \displaystyle{e=\sum\limits^\infty_{k=0}\frac{1}{k !}\Longrightarrow n!e=n!\left(1+1+\frac{1}{2!}+\ldots+\frac{1}{n!}+\ sum\limits^\infty_{k=n+1}\frac{1}{k!}\right)}$ $\displaystyle \Longrightarrow\displaystyle{\{n!e\}=\sum\limits_{ k=n+1}^\infty\frac{n!}{k!}}$

Tonio