Question

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

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

Answer 1

Answer: (B)

$\displaystyle\lim _{n \rightarrow \infty} n^{-n^{2}}\left[(n+1)\left(n+\frac{1}{2}\right) \ldots\left(n+\frac{1}{2^{n-1}}\right)\right]^{n}$
$=\displaystyle\lim _{n \rightarrow \infty}\left[\frac{(n+1)\left(n+\frac{1}{2}\right) \ldots\left(n+\frac{1}{2^{n-1}}\right)}{n^{n}}\right]^{n}$
$=\displaystyle\lim _{n \rightarrow \infty}\left(\frac{n+1}{n}\right)^{n} \cdot\left(\frac{n+\frac{1}{2}}{n}\right)^{n} \ldots\left(\frac{n+\frac{1}{2^{n-1}}}{n}\right)^{n}$
$=\displaystyle\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n} \cdot\left(1+\frac{1}{2 n}\right)^{n} \ldots\left(1+\frac{1}{2^{n-1} n}\right)^{n}\,\,\,\left(1^{\infty}\right.$ form $)$
$=\displaystyle\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n} \cdot\left(1+\frac{1}{2 n}\right)^{\frac{2 n}{2}} \ldots\left(1+\frac{1}{2^{n-1} n}\right)^{\frac{2^{n-1} \cdot n}{2^{n-1}}}$
$=e^{1} \cdot e^{1 / 2} \cdot e^{1 / 4} \ldots e^{1 / 2 n-1}$
$\ldots\left\{\right.$ using; $\left.\displaystyle\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{a n}=e^{a}\right\}$
$=e^{(1+1 / 2+1 / 4+\ldots)}=e^{\frac{1}{1-\frac{1}{2}}}=e^{2}$