Question

$\underset{n\to \infty }{lim}{\left(\frac{1^1\cdot 2^2\cdot 3^3.....(n-1{)}^{n-1}\cdot n^n}{n^{1+2+3+....+n}}\right)}^{\frac{1}{n^2}}$ equals.

Answer 1

Given: ${lim}_{n\to \infty }\frac{{(1^1.2^2.3^3......{(n-1)}^{n-1}.n^n)}^{\frac{1}{n^2}}}{n^{1+2+3+........n}}$

To find: Value of the limit

Sol: $l={lim}_{n\to \infty }{(\frac{1^1}{n^1}.\frac{2^2}{n^2}.\frac{3^3}{n^3}.......\frac{{(n-1)}^{n-1}}{n^{n-1}}.\frac{n^n}{n^n})}^{\frac{1}{n^2}}$

$⟹l={lim}_{n\to \infty }{\left(\right(\frac{1}{n}{)}^1.(\frac{2}{n}{)}^2.(\frac{3}{n}{)}^3........(\frac{n-1}{n}{)}^{n-1}.(\frac{n}{n}{)}^n)}^{\frac{1}{n^2}}$

Taking log on both sides

$ln\left(l\right)={lim}_{n\to \infty }{ln\left\{\right(\frac{1}{n}{)}^1.(\frac{2}{n}{)}^2.........(\frac{n}{n}{)}^n\}}^{\frac{1}{n^2}}$

$⟹ln\left(l\right)=\frac{1}{n^2}.{lim}_{n\to \infty }\left[ln\right(\frac{1}{n})+2ln(\frac{2}{n}).........nln(\frac{n}{n}\left)\right]$

$=\frac{1}{n^2}.{lim}_{n\to \infty }{\sum }_{r=1}^{n}r.ln\left(\frac{r}{n}\right)$

$=\frac{1}{n}.{lim}_{n\to \infty }{\sum }_{r=1}^n\frac{r}{n}.ln\left(\frac{r}{n}\right)$

$⟹ln\left(l\right)={\int }_0^{1}nln\left(n\right)dn$

Integrating by parts, we get

$ln\left(l\right)=ln\left(n\right).{\left[\frac{n^2}{2}\right]}_0^1-{\left[\frac{n^2}{4}\right]}_0^1$

$=0-\frac{1}{4}=\frac{-1}{4}$

$⟹l=e^{\frac{-1}{4}}$

Hence correct answer is $e^{\frac{-1}{4}}$