Question

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

• A. $e$
• B. $e^2$
• C. $e^4$
• D. None of these

Answer 1

Answer: (B)

$\underset{n\to \infty }{\lim }{n}^{-n^2}{\left[(n+1)\left(n+\frac{1}{2}\right)\dots \left(n+\frac{1}{2^{n-1}}\right)\right]}^n$
$=\underset{n\to \infty }{\lim }{\left[\frac{(n+1)\left(n+\frac{1}{2}\right)\dots \left(n+\frac{1}{2^{n-1}}\right)}{n^n}\right]}^n$
$=\underset{n\to \infty }{\lim }{\left(\frac{n+1}{n}\right)}^n\cdot {\left(\frac{n+\frac{1}{2}}{n}\right)}^n\dots {\left(\frac{n+\frac{1}{2^{n-1}}}{n}\right)}^n$
$=\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n\cdot {\left(1+\frac{1}{2n}\right)}^n\dots {\left(1+\frac{1}{2^{n-1}n}\right)}^n(1^{\infty }$ form $)$
$=\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n\cdot {\left(1+\frac{1}{2n}\right)}^{\frac{2n}{2}}\dots {\left(1+\frac{1}{2^{n-1}n}\right)}^{\frac{2^{n-1}\cdot n}{2^{n-1}}}$
$=e^1\cdot e^{1/2}\cdot e^{1/4}\dots e^{1/2n-1}$
$\dots \{$ using; $\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^{an}=e^a\}$
$=e^{(1+1/2+1/4+\dots )}=e^{\frac{1}{1-\frac{1}{2}}}=e^2$