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Question Number 167295 by mnjuly1970 last updated on 12/Mar/22

Answered by qaz last updated on 12/Mar/22

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\left(\frac{\left(2\mathrm{n}\right)!}{{\mathrm{n}}^{\mathrm{n}}\mathrm{n}!}\right)}^{1/\mathrm{n}}$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{\left(2\mathrm{n}\right)!}{{\mathrm{n}}^{\mathrm{n}}\mathrm{n}!}\cdot \frac{{(\mathrm{n}-1)}^{\mathrm{n}-1}(\mathrm{n}-1)!}{(2\mathrm{n}-2)!}$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{2(2\mathrm{n}-1)}{\mathrm{n}-1}{(1-\frac{1}{\mathrm{n}})}^{\mathrm{n}}$$$$={4\mathrm{e}}^{-1}$$

Answered by Mathspace last updated on 13/Mar/22

$$n!\sim n^n{e}^{-n}\sqrt{2\pi n}$$$$\left(2n\right)!\sim {\left(2n\right)}^{2n}{e}^{-2n}\sqrt{4\pi n}\Rightarrow$$$$\frac{\left(2n\right)!}{n^n.n!}=\frac{4^n{n}^{2n}{e}^{-2n}2\sqrt{\pi n}}{n^n{n}^n{e}^{-n}\sqrt{2\pi n}}$$$$=\sqrt{2}.4^n.e^{-n}\Rightarrow$$$${\left(\frac{\left(2n\right)!}{n^{n}n!}\right)}^{\frac{1}{n}}\sim {\left(\sqrt{2}\right)}^{\frac{1}{n}}.4.e^{-1}\to \frac{4}{e}\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{\left(\frac{\left(2n\right)!}{n^{n}n!}\right)}^{\frac{1}{n}}=\frac{4}{e}$$

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