$\lim\limits_{n\to\infty} \frac{n}{\sqrt[n]{n!}} =e$

Question

I don't know how to prove that $$\lim_{n\to\infty} \frac{n}{\sqrt[n]{n!}} =e.$$ Are there other different (nontrivial) nice limit that gives $e$ apart from this and the following $$\sum_{k = 0}^\infty \frac{1}{k!} = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e\;?$$

Answer 1

There are as many representations of $e$ as you could want at Wikipedia. $$e=\sum_1^{\infty}{k^7\over877k!}$$ $$e=\lim_{n\to\infty}n^{\pi(n)/n}$$ where $\pi(n)$ is the number of primes up to $n$. Any many, many more.

Answer 2

In the series for $$e^n=\sum_{k=0}^\infty \frac{n^k}{k!},$$ the $n$th and biggest(!) of the (throughout positve) summands is $\frac{n^n}{n!}$. On the other hand, all summands can be esimated as $$ \frac{n^k}{k!}\le \frac{n^n}{n!}$$ and especially those with $k\ge 2n$ can be estimated $$ \frac{n^k}{k!}<\frac{n^{k}}{(2n)^{k-2n}\cdot n^{n}\cdot n!}=\frac{n^{n}}{n!}\cdot \frac1{2^{k-2n}}$$ and thus we find $$\begin{align}\frac{n^n}{n!}<e^n&=\sum_{k=0}^{2n}\frac{n^k}{k!}+ \sum_{k=2n}^\infty \frac{n^k}{k!}\\&<(2n+1)\cdot\frac{n^n}{n!}+ \frac{n^n}{n!}\sum_{k=0}^\infty 2^{-k}\\&=(2n+3)\cdot\frac{n^n}{n!}.\end{align}$$ Taking $n$th roots we find $$ \frac n{\sqrt[n]{n!}}\le e\le \sqrt[n]{2n+3}\cdot\frac n{\sqrt[n]{n!}}.$$ Because $\sqrt[n]{2n+3}\to 1$ as $n\to \infty$, we obtain $$\lim_{n\to\infty}\frac n{\sqrt[n]{n!}}=e$$ from squeezing.

Answer 3

I will use Cauchy-D'Alembert criterion

$$\lim_{n\to\infty} \frac{n}{\sqrt[n]{n!}} =e$$

you can write the limit as $$\sqrt[n]{\frac{n^{n}}{n!}}.$$

Let be $\displaystyle x_{n}=\frac{n^{n}}{n!}.$

Now we can do $$\frac{x_{n+1}}{x_{n}}=\frac{\frac{(n+1)^{n}\cdot (n+1)}{n! \cdot (n+1)}}{\frac{n^n}{n!}}=\frac{(n+1)^{n}\cdot (n+1)}{n! \cdot (n+1)}\cdot \frac{n!}{n^{n}}=\frac{(n+1)^{n}}{n^n}=\left(\frac{n+1}{n}\right)^{n}=\left(1+\frac{1}{n}\right)^n \to e. $$

So $\displaystyle \lim_{n\to\infty} \frac{n}{\sqrt[n]{n!}} =e.$

Answer 4

Since $$ \ln \left( \frac{n}{\sqrt[n]{n!}}\right)=\ln n-\frac{\ln n!}{n} $$ all you need is the weak Stirling formula: $$ \ln n!=n\ln n -n +O(\ln n). $$ By comparison with the integral of the nondecreasing function $\ln t$: $$ \int_1^n\ln tdt\leq \ln n! =\sum_{k=2}^n \ln k\leq \int_2^{n+1}\ln t dt. $$ Recall that $\int \ln tdt =t\ln t -t +C$ and compute the lhs and the rhs.

The weak Stirling formula follows easily.

Answer 5

Use Stirling's approximation

$$\frac{n}{\sqrt[n]{n!}}\cong\frac{n}{\sqrt[n]{\sqrt{2\pi }\,n^{n+1/2}e^{-n}}}=\frac{1}{(2\pi)^{1/2n}}\frac{e}{n^{1/2n}}\xrightarrow[n\to\infty]{}1\cdot\frac{e}{1}=e$$

Answer 6

I found this here:

$$e = \lim_{n \to \infty} \sqrt[n]{n\#}$$

Also see this related older post.