{en!} ...fractional part

[sasa]

Suppose that $\{\cdot \}$ denotes fractional part and $e$ is Napier'constant. If $a=1 \; ,\; b=e-1\; ,$ show that


$ \begin{array}{|l|} \hline \\ \displaystyle \frac{1}{n+1}+\frac{e^{\frac{a}{n+3}}}{(n+1)(n+2)} < \displaystyle \{en!\}=e\int\limits_{0}^1 e^{-t}t^n\; dt <\displaystyle \frac{1}{n+1}+\frac{e^{\frac{b}{n+3}}}{(n+1)(n+2)} \\ \\ \hline \end{array} . $

See also Mehdi Hassani , Derangements and Applications , Journal of Integer Sequences, vol.6 (2003), article 03.1.2

[Руст]

Пусть $y(n)=\{en!\} =\frac {1}{n+1}+\frac{1}{(n+1)(n+2)}+\dots=\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}x(n)$, где
$ x(n)=1+\frac{1}{n+3}+\frac{1}{(n+3)(n+4)}+\dots $. Так как (n+3+k)<(k+1)(n+3), получаем нижнюю оценку. Чуть посложнее получается верхня оценка для x(n).

[maxal]

http://www.mathlinks.ro/Forum/viewtopic.php?t=78473

[sasa]

maxal писал(а):

http://www.mathlinks.ro/Forum/viewtopic.php?t=78473

Ok, but until now ...no solution ...or no interest ./sasa