Sunday, August 4, 2019

real analysis - Show that $limlimits_{n to infty} frac{(n!)^{1/n}}{n}= frac{1}{e}$






Show that $$\lim_{n \to \infty} \left\{\frac{(n!)^{1/n}}{n}\right\} = \frac{1}{e}$$




What I did is to let $U_n = \dfrac{(n!)^{\frac{1}{n}}}{n}$ and $U_{n+1} = \dfrac{(n+1)!^{\frac{1}{n+1}}}{n+1}$. Then




$$\frac{ U_{n+1} }{U_n } = \frac{\frac{(n+1)!^{\frac{1}{n+1}}}{n+1}}{\frac{(n!)^{\frac{1}{n}}}{n}}$$



Next I just got stuck. Am I on the right track, or am I wrong doing this type of sequence?


Answer



Let $v_n = \frac{n!}{n^n } $ then $$ \frac{v_{n+1}}{v_n } =\frac{(n+1)! }{(n+1)^{n+1}} \cdot \frac{n^n }{n!} =\frac{n^n}{(n+1)^n }=\frac{1}{\left(1+\frac{1}{n}\right)^n}\to \frac{1}{e}$$ hence $$\frac{\sqrt[n]{n!} }{n} =\sqrt[n]{v_n} \to\frac{1}{e} .$$


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