Calculus

Used that if $\lim_{x\rightarrow n}$ $f(x)^{g(x)}$ = $1^{\infty}$ then req limit is $e^{a}$ where a is $(f(x)-1)\times g(x)$

Since $\sum _{k=1}^nk^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ and $\sum _{k=1}^nk=\frac{n\left(n+1\right)}{2}$ , $\displaystyle \large \lim_{n \to \infty} \left[\frac{\left[\frac{2n\left(n\right)\left(n+1\right)}{2}\right]}{\left[\frac{3\left(n\right)\left(n+1\right)\left(2n+1\right)}{6}\right]}\right]^n$ $=\displaystyle \large \lim_{n \to \infty} \left(\frac{2n}{2n+1}\right)^n$

$=\displaystyle \large \lim_{n \to \infty} \left(1-\frac{1}{2n+1}\right)^n$

$=\displaystyle \large \lim_{n \to \infty} \left(1+\frac{1}{-\left(2n+1\right)}\right)^n$

Now, let $y=-2n-1$ , then:

$=\displaystyle \large \lim_{y \to \infty} \left(1+\frac{1}{y}\right)^{-\frac{\left(y+1\right)}{2}}$

$=\displaystyle \large \lim_{y \to \infty} \left(1+\frac{1}{y}\right)^{-\frac{y}{2}}\cdot \left(1+\frac{1}{y}\right)^{-\frac{1}{2}}$

$=\displaystyle \large \lim_{y \to \infty} \left[\left(1+\frac{1}{y}\right)^y\right]^{-\frac{1}{2}}\cdot \left(1+0\right)^{-\frac{1}{2}}$

Since $=\displaystyle \large \lim_{x \to \infty} \left(1+\frac{1}{x}\right)^x=e$ ,

$\displaystyle N = \large e^{-\frac{1}{2}}=\frac{1}{\sqrt{e}}$