<s>[INST] Problem with inequality and number e Prove that, for every positive integer $n$, the following inequality holds:
$$n \{n! \times e\} \lt 1,$$ where $\{x\}$ denotes the fractional part function applied to number $x$.
I can;t think of any solution.
 [/INST] Because 
$$e=2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+...=$$
$$=2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+\frac{1}{(n+1)!}\left(1+\frac{1}{n+2}+\frac{1}{(n+2)(n+3)}+...\right)<$$
$$=2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+\frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+...\right)=$$
$$=2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+\frac{1}{(n+1)!}\cdot\frac{1}{1-\frac{1}{n+1}}=$$
$$=2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+\frac{1}{n\cdot n!}.$$
Thus, $\{en!\}<\frac{1}{n}$ and we are done!
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