Question

The value of
$\lim \: n \: to \: \infty \: { (\frac{n!}{ {n}^{n} } )}^{ \frac{1}{n} } \: is \:$
Please provide the detailed solution.

Urgently required

Moderators, math genius, Brainly Stars ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty } \: {\bigg[\dfrac{n!}{ {n}^{n} } \bigg]}^{ \dfrac{1}{n} }$

Let assume that,

$\rm :\longmapsto\:y = \displaystyle\lim_{n \to \infty } \: {\bigg[\dfrac{n!}{ {n}^{n} } \bigg]}^{ \dfrac{1}{n} }$

On taking log on both sides, we get

$\rm :\longmapsto\:logy = \displaystyle\lim_{n \to \infty } \: log {\bigg[\dfrac{n!}{ {n}^{n} } \bigg]}^{ \dfrac{1}{n} }$

We know,

$\rm :\longmapsto\:logy = \displaystyle\lim_{n \to \infty } \dfrac{1}{n}\: log {\bigg[\dfrac{n!}{ {n}^{n} } \bigg]}$

$\rm :\longmapsto\:logy = \displaystyle\lim_{n \to \infty } \dfrac{1}{n}\: log {\bigg[\dfrac{n(n - 1)(n - 2) - - - 1}{ {n}.n.n - - - .n} \bigg]}$

$\rm :\longmapsto\:logy = \displaystyle\lim_{n \to \infty } \dfrac{1}{n}\: log {\bigg[\dfrac{n(n - 1)(n - 2) - - -[n - (n - 1)]}{ {n}.n.n - - - .n} \bigg]}$

$\rm :\longmapsto\:logy = \displaystyle\lim_{n \to \infty } \dfrac{1}{n}\: \sum_{r=0}^{n-1}log\bigg[\dfrac{n - r}{n} \bigg]$

$\rm :\longmapsto\:logy = \displaystyle\lim_{n \to \infty } \dfrac{1}{n}\: \sum_{r=0}^{n-1}log\bigg[1 - \dfrac{r}{n} \bigg]$

Now, By using limit as a sum

$\rm :\longmapsto\:logy = \displaystyle\int_{0}^{1} \: log(1 - x) \: dx$

Now, we know,

$\boxed{ \bf{ \: \displaystyle\int_{0}^{a} \: f(x) \: dx \: = \: \displaystyle\int_{0}^{a} \: f(a - x) \: dx}}$

So, using this identity, we get

$\rm :\longmapsto\:logy = \displaystyle\int_{0}^{1} \: log[1 - (1 - x)] \: dx$

$\rm :\longmapsto\:logy = \displaystyle\int_{0}^{1} \: log[1 - 1 + x] \: dx$

$\rm :\longmapsto\:logy = \displaystyle\int_{0}^{1} \: logx \: dx$

can be rewritten as

$\rm :\longmapsto\:logy = \displaystyle\int_{0}^{1} \: 1.logx \: dx - - - (1)$

Consider,

$\red{\rm :\longmapsto\:\displaystyle\int1.logxdx}$

Now, using By parts, we get

$\rm \: = \: logx \displaystyle\int \: 1\: dx - \displaystyle\int\bigg[\dfrac{d}{dx}logx\displaystyle\int1dx\bigg]dx$

$\rm \: = \:logx \: (x) - \displaystyle\int \: \dfrac{1}{x} \times x \: dx$

$\rm \: = \:logx \: (x) - \displaystyle\int \: 1 \: dx$

$\rm \: = \:x \: logx \: - \: x$

$\red{\bf\implies \:\displaystyle\int \: logx \: dx \: = \: x \: logx \: - \: x}$

On substituting this value in equation (1), we get

$\rm :\longmapsto\:logy = \bigg[x \: logx \: - \: x \bigg]_{0}^{1}$

$\rm :\longmapsto\:logy = \bigg[1 \: log1 \: - \: 1 \bigg] - 0$

$\rm :\longmapsto\:logy = \bigg[1 \times 0 \: - \: 1 \bigg]$

$\rm :\longmapsto\:logy = - 1$

$\bf\implies \:y = {e}^{ - 1}$

Hence,

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty } \bf\: {\bigg[\dfrac{n!}{ {n}^{n} } \bigg]}^{ \dfrac{1}{n} } = \dfrac{1}{e}$