Question

$\underset{n\to \infty }{lim}{\left(\frac{n!}{(mn{)}^n}\right)}^{1/n}\left(mϵN\right)$ is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$\underset{h\to \infty }{lim}{\left(\frac{n!}{(mn{)}^n}\right)}^{\frac{1}{n}}=\underset{h\to \infty }{lim}{\left(\frac{\mathrm{1.2.3.4}\dots \dots (n-1)n}{n^n}\right)}^{\frac{1}{n}}\times \frac{1}{m}P=\underset{h\to \infty }{lim}\left(\frac{1}{n}\right)\left(\frac{2}{n}\right)\left(\frac{3}{n}\right)\dots \dots \left(\frac{n-1}{n}\right){\left(\frac{n}{n}\right)}^{\frac{1}{n}}\times \frac{1}{m}lo{g}_{e}P=\underset{n\to \infty }{lim}\frac{1}{n}{\Sigma }_{r=1}^{n}lo{g}_e\left(\frac{r}{n}\right)+lo{g}_e\frac{1}{m}={\int }_0^{1}lo{g}_{e}xdx-lo{g}_{e}m=-1-lo{g}_{e}mlo{g}_{e}P=-lo{g}_{e}e-lo{g}_{e}m=-lo{g}_e\left(em\right)=lo{g}_e\left(\frac{1}{em}\right)\therefore P=\frac{1}{em}$