Question

chapter - definite integral​

Answer 1

Let,

$\displaystyle\longrightarrow y=\lim_{n\to\infty}\left(\dfrac{n!}{n^n}\right)^{\dfrac{1}{n}}$

Since $n!=1\cdot2\cdot3\cdot\,\dots\,\cdot n$ and $n^n=\underbrace{n\cdot n\cdot n\cdot\,\dots\,\cdot n}_{n\ times}$

$\displaystyle\longrightarrow y=\lim_{n\to\infty}\left(\dfrac{1\cdot2\cdot3\cdot\,\dots\,\cdot n}{\underbrace{n\cdot n\cdot n\cdot\,\dots\,\cdot n}_{n\ times}}\right)^{\dfrac{1}{n}}$

or,

$\displaystyle\longrightarrow y=\lim_{n\to\infty}\left(\dfrac{1}{n}\cdot\dfrac{2}{n}\cdot\dfrac{3}{n}\cdot\,\dots\,\cdot\dfrac{n}{n}\right)^{\dfrac{1}{n}}$

or,

$\displaystyle\longrightarrow y=\lim_{n\to\infty}\left(\prod_{r=1}^n\dfrac{r}{n}\right)^{\dfrac{1}{n}}$

Taking log on both sides,

$\displaystyle\longrightarrow\log y=\lim_{n\to\infty}\dfrac{1}{n}\log\left(\prod_{r=1}^n\dfrac{r}{n}\right)$

Since $\displaystyle\log\left(\prod_{r=a}^bx_r\right)=\sum_{r=a}^b\log(x_r),$

$\displaystyle\longrightarrow\log y=\lim_{n\to\infty}\dfrac{1}{n}\sum_{r=1}^n\log\left(\dfrac{r}{n}\right)\quad\quad\dots(1)$

We know the limit of a sum [or Riemann Sum] can be converted to a definite  integral as,

$\boxed{\displaystyle\lim_{n\to\infty}\dfrac{1}{n}\sum_{r=1}^nf\left(\dfrac{r}{n}\right)=\int\limits_0^1f(x)\ dx}$

In (1), $f\left(\dfrac{r}{n}\right)=\log\left(\dfrac{r}{n}\right)$ or $f(x)=\log x.$

Hence (1) becomes,

$\displaystyle\longrightarrow\log y=\int\limits_0^1\log x\ dx$

$\displaystyle\longrightarrow\log y=\Big[x\log x-x\Big]_0^1$

$\displaystyle\longrightarrow\log y=1\cdot\log 1-1-\lim_{x\to0}x\log x+0$

$\displaystyle\longrightarrow\log y=0-1-\lim_{x\to0}\dfrac{\log x}{\left(\dfrac{1}{x}\right)}$

Applying L'hospital's Rule in the limit,

$\displaystyle\longrightarrow\log y=-1-\lim_{x\to0}\dfrac{\left(\dfrac{1}{x}\right)}{\left(-\dfrac{1}{x^2}\right)}$

$\displaystyle\longrightarrow\log y=-1+\lim_{x\to0}x$

$\displaystyle\longrightarrow\log y=-1+0$

$\displaystyle\longrightarrow\log y=-1$

$\displaystyle\longrightarrow y=e^{-1}$

$\displaystyle\longrightarrow y=\dfrac{1}{e}$

Hence,

$\displaystyle\longrightarrow\underline{\underline{\lim_{n\to\infty}\left(\dfrac{n!}{n^n}\right)^{\dfrac{1}{n}}=\dfrac{1}{e}}}$