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Question Number 142464 by som(math1967) last updated on 01/Jun/21

$$\underset{\mathit{n}\to \mathrm{\infty }}{\mathit{l}\mathit{i}\mathit{m}}{\left[\frac{\left(2\mathit{n}\right)!}{\mathit{n}!{\mathit{n}}^{\mathit{n}}}\right]}^{\frac{1}{\mathit{n}}}=?$$

Answered by Dwaipayan Shikari last updated on 01/Jun/21

$$\underset{n\to \mathrm{\infty }}{\lim }{\left[\frac{2^{2n}{n}^{2n}{e}^{-2n}\sqrt{4\pi n}}{n^n{e}^{-n}\sqrt{2\pi n}{n}^n}\right]}^{\frac{1}{n}}=y$$$$\Rightarrow {\left[{\left(\frac{4}{e}\right)}^n\sqrt{2}\right]}^{\frac{1}{n}}=y\Rightarrow y=\frac{4}{e}$$

Answered by Dwaipayan Shikari last updated on 01/Jun/21

$$\frac{1}{n}log(\left(2n\right)!)-\frac{1}{n}log\left(\frac{n!}{n^n}\right)-\frac{1}{n}log\left(n^{2n}\right)=log\left(y\right)$$$$\frac{1}{n}log\left(\frac{\left(2n\right)!}{n^{2n}}\right)-\frac{1}{n}\underset{r=1}{\sum ^n}log\left(\frac{r}{n}\right)=log\left(y\right)$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{r=1}{\sum ^{2n}}log\left(\frac{r}{n}\right)-\frac{1}{n}\underset{r=1}{\sum ^n}log\left(\frac{r}{n}\right)=log\left(y\right)$$$$\Rightarrow {\int }_0^{2}log\left(x\right)dx-{\int }_0^{1}log\left(x\right)dx=log\left(y\right)$$$$\Rightarrow log\left(4\right)-2+1=log\left(y\right)\Rightarrow y=\frac{4}{e}$$

Commented by som(math1967) last updated on 01/Jun/21

$$\mathit{T}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{y}\mathit{o}\mathit{u}$$

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