IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

neeraj_agarwal_1990

$\mathop{\lim}\limits_{{n} \to {\infty}}{\frac{n!}{{mn}^{n}}}^{1/n}$ =?

(1/n is exponent of the whole term)

ans- $({em})^{-1}$

ramkumar_november

l = $\lim_{n\to\infty}\frac{(n!)^{\frac{1}{n}}}{m.n}$

According to stirling's approximation,

$(n!)^{\frac{1}{n}}=\frac{n}{e}$

$\frac{(n!)^{\frac{1}{n}}}{n}=\frac{1}{e}$

so the limit becomes

l = $\lim_{n\to\infty}\frac{1}{e.m}$

l = $(e.m)^{-1}$

neeraj_agarwal_1990

i don't know abt that approximation...is there any other method?

studyid

One Aliter :

$\mathop{\lim}\limits_{{n} \to {\infty}} \frac{n!}{m{n}^{n}}^{\frac{1}{n}}$

$=\frac{1}{m} \mathop{\lim}\limits_{{n} \to {\infty}}\frac{n!}{{n}^{n}}^{\frac{1}{n}}$

Now let

l = $\mathop{\lim}\limits_{{n} \to {\infty}}\frac{n!}{{n}^{n}}^{\frac{1}{n}}$

$log l = \mathop{\lim}\limits_{{n} \to {\infty}}\frac{1}{n}[log(\frac{n!}{{n}^{n}})]$

$= \mathop{\lim}\limits_{{n} \to {\infty}}\frac{1}{n}[log(\frac{n(n-1)(n-2)....(n-(n-1))}{n.n.n.n.n..............n\;times})]$

$=\mathop{\lim}\limits_{{n} \to {\infty}} \frac{1}{n}log[(\frac{n}{n})(\frac{n-1}{n})...(\frac{2}{n})(\frac{1}{n})]$

$=\mathop{\lim}\limits_{{n} \to {\infty}}(\frac{1}{n})\sum_{r=0}^{n-1}log[\frac{n-r}{n}]$

$= \int_{0}^{1}log (1-x)dx$

There fore l = $\frac{1}{e}$

And hence the required limit as asked in the question is $\frac{1}{e.m}$

= ${(em)}^{-1}$