# An interesting limit

• Nov 22nd 2010, 11:44 AM
An interesting limit
$\lim \{n!e\}= \:?$
where {.} is the symbol of the fractinal part

• Nov 22nd 2010, 12:00 PM
roninpro
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.
• Nov 22nd 2010, 12:04 PM
tonio
Quote:

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