Question

If $L=\underset{n\to \infty }{lim}{n}^{-n^2}{[(n+1)(n+\frac{1}{2})(n+\frac{1}{2^2})\cdots (n+\frac{1}{2^{n-1}})]}^n$, then the value of $\ln L$ is

Answer 1

$\underset{n\to \infty }{lim}{n}^{-n^2}{[(n+1)(n+\frac{1}{2})(n+\frac{1}{2^2})\cdots (n+\frac{1}{2^{n-1}})]}^n$

$=\underset{n\to \infty }{lim}{\left[\frac{(n+1)(n+\frac{1}{2})(n+\frac{1}{2^2})\cdots (n+\frac{1}{2^{n-1}})}{n^n}\right]}^n$

$=\underset{n\to \infty }{lim}[{(1+\frac{1}{n})}^n{(1+\frac{1}{2n})}^n{(1+\frac{1}{2^{2}n})}^n\cdots {(1+\frac{1}{2^{n-1}n})}^n]$

$=e^1\cdot e^{1/2}\cdot e^{1/2^2}\cdot \cdots e^{1/2^{n-1}}\cdots \text{upto}\infty$

$=e^{1+1/2+1/2^2+\cdots }$

$=e^2$