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Question Number 121716 by bemath last updated on 11/Nov/20

$$\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{(n+1)(n+2)(n+3)...\left(3n\right)}{n^{2n}}\right)}^{\frac{1}{n}}=?$$

Answered by Dwaipayan Shikari last updated on 11/Nov/20

$$\underset{n\to \mathrm{\infty }}{\lim }{((1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{2n}{n}))}^{\frac{1}{n}}=y$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{r=1}{\sum ^{2n}}log(1+\frac{r}{n})=logy$$$${\int }_0^{2}log(1+x)dx=logy$$$${[xlog(1+x)-x+log(x+1)]}_0^2=logy$$$$3log3-2=logy$$$$y=\frac{27}{e^2}$$

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