Proving $\lim_{ x \rightarrow \infty} x/(x!)^{1/x} = e$ without Stirling?

Question

Show that:

$\lim_{ x \rightarrow \infty} x/(x!)^{1/x} = e$

I could prove this by using Stirling’s approximation. But I was wondering if there’s a more elementary way of doing it.

Answer 1

There's Cauchy's limit theorem which states that if $a_{n}$ is a sequence of positive real numbers such that $\lim\frac{a_{n+1}}{a_{n}}=l<\infty$ exists then $\lim_{n\to\infty}(a_{n})^{\frac{1}{{n}}}=l$ . Note that this should remind you of the ratio test and the root test for convergence of series .

Hence consider the sequence $\bigg(\frac{n^{n}}{n!}\bigg)^{\frac{1}{n}}$ and try to apply the above theorem. You should get $e$ as the limit of the ratio of $n+1$ th and the $n$ th term.

Alternatively :-

Consider $a_{n}=\bigg(\prod_{r=1}^{n}\frac{n}{r}\bigg)^{\frac{1}{n}}$ . So taking logs on both sides, you get:-

$$\log(a_{n})=\frac{1}{n}\sum_{r=1}^{n}\log(\frac{n}{r})$$

Using Riemann Sums, you have:-

$\lim_{n\to\infty}\log(a_{n})=\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}\log(\frac{n}{r})=\int_{0}^{1}-\ln(x)\,dx = 1$

Thus $\lim_{n\to\infty}\log(a_{n})= 1\implies\log(\lim_{n\to\infty}a_{n})=1$ by continuity .

Hence , $\lim_{n\to\infty}a_{n} = e $ . (Notice that I am apriori assuming the limit of $a_{n}$ exists and just finding the value of it. I have not shown or proven existence anywhere). So technically, to justify everything, one should first prove that $a_{n}$ is indeed a Cauchy Sequence.

Answer 2

Another solution is based on the MacLaurin expansion $$e^x=\sum_{k=0}^\infty {x^k\over k!}$$ We have $$e^n=\sum_{k=0}^\infty {n^k\over k!}\ge {n^n\over n!}$$ The sequence $a_k= {k^n\over k!}$ is increasing for $0\le k\le n,$ decreasing for $k\ge n$ and $${a_{k+1}\over a_k}={n\over k+1}\le {n\over n+1},\quad k\ge n$$ Therefore $$e^n\le (n+1){n^n\over n!}+\sum_{k=n}^\infty \left ({n\over n+1}\right )^{k-n}\, {n^n\over n!}\\ = (n+1){n^n\over n!}+ (n+1){n^n\over n!}=2(n+1){n^n\over n!}$$ Therefore $${e^n\over 2(n+1)}\le {n^n\over n!}\le e^n$$ The conclusion follows by the squeezing principle.