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I'd love your help this time with the following limit:

$\lim_{n \to +\infty } \left \{ en! \right \}$

when $\{ a \}=a-[a].$

Honestly, I don't have a clue.

Thank you.

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    $\begingroup$ You're probably familiar with at least two expressions for $e$. Which one seems relevant? What happens when you use it? $\endgroup$ – Alon Amit Jun 19 '11 at 7:47
  • $\begingroup$ @Alon: I don't get you. What two expressions? $\endgroup$ – user6163 Jun 19 '11 at 7:54
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    $\begingroup$ @Nir: There are more than two here. Alon Amit is referring to the two most famous, each sometimes given as the definition of $e$, and one (which appears on the linked page) is especially helpful for this problem. $\endgroup$ – Jonas Meyer Jun 19 '11 at 7:59
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    $\begingroup$ @Alon,@Jonas: Do you mean $e=\sum_{k=0}^{\infty}\frac{1}{k!}$? $\endgroup$ – user6163 Jun 19 '11 at 8:03
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You have $n!e=n!(1+\frac{1}{2!}+...+\frac{1}{n!})+\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+....+\frac{1}{(n+1)(n+2)....(n+p)}+...$.

We have $\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+....+\frac{1}{(n+1)(n+2)....(n+p)}+...$ $\leq \frac{1}{n+1}+\frac{1}{(n+1)^2}+...+\frac{1}{(n+1)^p}+...=$ $=\frac{1}{n+1}(1+\frac{1}{n+1}+...+\frac{1}{(n+1)^p}+..)=\frac{1}{n+1}\frac{1}{1-\frac{1}{n+1}}=\frac{1}{n}$.

This means that $\{n!e\}\leq \frac{1}{n}$, and the conclusion follows.

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Use the fact that $$ en! = n! \cdot\biggl(1 + \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{n!} \biggr) + n! \cdot \biggl( \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots \biggr)$$

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