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

Question

Aicharya · Accepted Answer

Let $P=\lim _{n \rightarrow \infty} \frac{1}{n}[(n+1)(n+2) \ldots \ldots(2 n)]^{1 / n}$ On applying ' $\log ^{\prime}$ both sides, we get $\log P=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\log \left(\frac{n+1}{n}\right)\left(\frac{n+2}{n}\right) \ldots \ldots\left(\frac{n+n}{n}\right)\right]$ $\begin{aligned} & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1+\frac{r}{n}\right)=\int_0^1 \log (1+x) d x \ & =[x \log (1+x)]_0^1-\int_0^1 \frac{x}{1+x} d x \ & =\log 2-(x-\log (1+x))_0^1 \ & =\log 2-1+\log 2 \ & =\log 4-1=\log \left(\frac{4}{e}\right) \Rightarrow P=\frac{4}{e} \end{aligned}$ Hence, option (d) is correct.

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

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