Question

If the value of $\underset{n\to \infty }{lim}{\left(\frac{(2n+1)!}{n^{2n+1}}\right)}^{\frac{1}{n}}=\ln a+b,$ then the value of $(a-b)$ equals to (where $a,b\in Z$)

Answer 1

$L=\underset{n\to \infty }{lim}{\left(\frac{(2n+1)!}{n^{2n+1}}\right)}^{\frac{1}{n}}=\underset{n\to \infty }{lim}{\left(\frac{1\cdot 2\cdot 3\cdots (2n+1)}{n^{2n+1}}\right)}^{\frac{1}{n}}$
taking $\ln$ both sides,
$\Rightarrow \ln L=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^{2n+1}\ln \left(\frac{r}{n}\right)=\underset{0}{\overset{2}{\int }}\ln xdx=\left[x\right(\ln x-1){]}_0^2=\ln 4-2$
$\Rightarrow a=4,b=-2$
So, $a-b=6$