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Find the value of $\lim_{n \to \infty} \frac{(n!)^{\frac{1}{n}}}{n}$

I found a solution here which says: $\lim _{n \rightarrow \infty} \frac{(n !)^{\frac{1}{n}}}{n}=\lim _{n \rightarrow \infty} \sqrt[n]{\frac{n !}{n^n}}=\lim _{n \rightarrow \infty} \sqrt[n]{a_n}=\lim _{n \rightarrow \infty} \frac{a_{n+1}}{a_n}=\lim _{n \rightarrow \infty} \frac{1}{\left(1+\frac{1}{n}\right)^n}=\frac{1}{e}$ My query is how they got $\lim _{n \rightarrow \infty} \sqrt[n]{a_n}=\lim _{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$. I get that they took $a_n = n!/n^n$, but how does that correspond to $\lim _{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$?

zaemon_23
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1 Answers1

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Use $\log$: $$\lim_{n \to \infty} \log\left(\frac{(n!)^{\frac{1}{n}}}{n}\right) = \lim_{n \to \infty} \log\left(\frac{n!}{n^n}\right)^{\frac{1}{n}} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}\log\left(\frac{k}{n}\right) = \int_{0}^{1} \log(x) dx \\ = [x\log(x)]_{x=0}^{x=1}-\int_{0}^{1} 1 dx = -1$$ Thus: $$\lim_{n \to \infty} \frac{(n!)^{\frac{1}{n}}}{n} = \exp\left( \lim_{n \to \infty} \log\left(\frac{(n!)^{\frac{1}{n}}}{n}\right)\right) = \exp(-1) = \frac{1}{e}$$

psl2Z
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