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.
2026-03-25 18:52:30.1774464750
Problem with inequality and number e
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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!