lim _{n rightarrow infty} frac{1}{n} [(n+1)(n | vedclass

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Question

(D) Let $P = \lim _{n \rightarrow \infty} \frac{1}{n} [(n+1)(n+2) \cdots (2n)]^{\frac{1}{n}}$.Taking the natural logarithm on both sides:$\log P = \li

Vedclass · Accepted Answer

$\frac{4}{e}$

Vedclass · Answer

(D) Let $P = \lim _{n \rightarrow \infty} \frac{1}{n} [(n+1)(n+2) \cdots (2n)]^{\frac{1}{n}}$.Taking the natural logarithm on both sides:$\log P = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \log \left( \frac{n+r}{n} \right) = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \log \left( 1 + \frac{r}{n} \right)$.This is a Riemann sum,which can be expressed as the definite integral:$\int_{0}^{1} \log(1+x) dx$.Using integration by parts,$\int \log(1+x) dx = (1+x)\log(1+x) - (1+x) + C$.Evaluating from $0$ to $1$:$[(1+x)\log(1+x) - (1+x)]_{0}^{1} = (2\log 2 - 2) - (0 - 1) = 2\log 2 - 1 = \log 4 - \log e = \log \left( \frac{4}{e} \right)$.Since $\log P = \log \left( \frac{4}{e} \right)$,we have $P = \frac{4}{e}$.Thus,option $(D)$ is correct.

$\lim _{n \rightarrow \infty} \frac{1}{n} [(n+1)(n+2) \cdots (2n)]^{\frac{1}{n}} = $

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