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Possible Duplicate:
Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$

Stirling's approximation says that $$ \lim_{n \to \infty} \frac{n^n \sqrt{n}}{n! e^n } = \frac{1}{\sqrt{2 \pi}}.$$ Since $\lim_{n \to \infty} x^\frac{1}{n} \to 1$ uniformly on a neighbourhood of $\frac{1}{\sqrt{ 2 \pi}}$, it follows that $$\lim_{n \to \infty} \left( \frac{n^n \sqrt{n}}{n! e^n } \right)^\frac{1}{n} = \lim_{n \to \infty} \frac{n}{(n!)^\frac{1}{n}} \cdot \frac{n^\frac{1}{2n}}{e} = 1.$$ Since $\lim_{n \to \infty} n^\frac{1}{2n} = 1$, we get the limit in the title $$\lim_{n \to \infty} \frac{n}{(n!)^\frac{1}{n}} = e.$$

Question: Is Stirling's approximation is really needed to derive the above limit? Or is there an easier way to reach the same conclusion?

Motivation: The radius of convergence $R$ of a power series $\sum_{n=0}^\infty a_nx^n$ is given by Hadamard's formula $$\frac{1}{R} = \limsup_{n \to \infty} |a_n|^\frac{1}{n}.$$ If we know ahead of time that $R > 0$ then the coefficients are given by $$ a_n = \frac{f^{(n)}(0)}{n!}$$ where $f$ is the function defined by the power series. Then we get $$\frac{1}{R} = \limsup_{n \to \infty} |a_n|^\frac{1}{n} = \limsup_{n \to \infty} \frac{|f^{(n)}(0)|^\frac{1}{n}}{n} \cdot \frac{n}{(n!)^\frac{1}{n}} = e \cdot \limsup_{n \to \infty} \frac{|f^{(n)}(0)|^\frac{1}{n}}{n}.$$ So one use of the limit is to clean up the formula for the radius of convergence of a power series in terms of the derivatives of the corresponding function.

Community
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Mike F
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    Similar questions have been discussed for a long time, and for example you can find an instance at here. – Sangchul Lee Nov 16 '12 at 07:27
  • You only need Stirling's approximation up to a polynomial multiplicative constant, which you can get from a straightforward Riemann sum argument. – Qiaochu Yuan Nov 16 '12 at 08:20

2 Answers2

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Taking the natural logarithm, you want to show that $$ \lim_{n\to\infty}\left(\ln n-\frac1{n}\sum_{i=1}^n\ln i\right)=1 $$ or equivalently $$ n \ln n - \sum_{i=1}^n\ln i \sim n\qquad\text{as }n\to\infty. $$ Now $$ \int_1^n\ln x\,dx<\sum_{i=1}^n\ln i<\int_1^n\ln x\,dx+\ln n, $$ and since $x\ln x-x$ is a primitive of $\ln x$, it follows that $$ \sum_{i=1}^n\ln i = n \ln n-n+O(\ln n) $$ which approximation suffices easily for the asymptotic equivalence above.

Marc van Leeuwen
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  • an alternative from the $2$nd line is: $\lim_{n\to\infty}\left(\ln n-\frac1{n}\sum_{i=1}^n\ln i\right)=\lim_{n\to\infty}\left(\ln n-\left(\frac1{n}\sum_{i=1}^n\ln \frac{i}{n}+\ln n\right)\right)=-\int_0^1 \ln x=1.$ – user 1591719 Nov 16 '12 at 11:53
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HINT: write $\lim_{n \to \infty} \displaystyle\frac{n}{(n!)^\frac{1}{n}}$ as $\lim_{n \to \infty} \left(\displaystyle\frac{n^n}{(n!)}\right)^\frac{1}{n}$ and then compute $\lim_{n \to \infty} \frac{a_{n+1}}{a_{n}}$ where $a_n=\displaystyle\frac{n^n}{(n!)}$

user 1591719
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