lim _{n} {rightarrow infty} n^{-n k} left{(n+ | vedclass

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(B) Let $P = \lim _{n}$ ${\rightarrow \infty} n^{-n k} \left\{(n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^2}\right) \ldots\left(n+\frac{1}{2^{k-

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$e^{2\left(1-\frac{1}{2^k}\right)}$

Vedclass · Answer

(B) Let $P = \lim _{n}$ ${\rightarrow \infty} n^{-n k} \left\{(n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^2}\right) \ldots\left(n+\frac{1}{2^{k-1}}\right)\right\}^n$. Taking the natural logarithm on both sides: $\log P = \lim _{n \rightarrow \infty} n \left[ \sum_{j=0}^{k-1} \log \left(1 + \frac{1}{2^j n}\right) \right]$. Using the standard limit $\lim _{x \rightarrow 0} \frac{\log(1+ax)}{x} = a$,we have: $\log P = \sum_{j=0}^{k-1} \lim _{n}$ ${\rightarrow \infty} \frac{\log \left(1 + \frac{1}{2^j n}\right)}{1/n} = \sum_{j=0}^{k-1} \frac{1}{2^j}$. This is a geometric series with $k$ terms,first term $a=1$ and common ratio $r=1/2$: $\sum_{j=0}^{k-1} \left(\frac{1}{2}\right)^j = \frac{1(1-(1/2)^k)}{1-1/2} = 2 \left(1 - \frac{1}{2^k}\right)$. Therefore,$P = e^{2 \left(1 - \frac{1}{2^k}\right)}$.

$\lim _{n}$ ${\rightarrow \infty} n^{-n k} \left\{(n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^2}\right) \ldots\left(n+\frac{1}{2^{k-1}}\right)\right\}^n=$

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