Find $\lim\limits_{n\to\infty} (n^n/n!)^{1/n}$ 
Find $\lim\limits_{n\to\infty} (n^n/n!)^{1/n}$.

One can use Stirling's approximation, which states that for all $n\ge 1, n! = \sqrt{2\pi n} (\frac{n}e)^{n}e^{\frac{\theta_n}{12n}}$, where $\theta_n \in(0,1).$
Using this approximation yields the required limit of $e$ using the fact that $\lim\limits_{n\to\infty} n^{1/(2n)} = 1.$
But isn't there a simpler approach that involves using the root test? For instance, it might be useful to observe that the radius of convergence of the series $$\sum_{n=1}^\infty \frac{n!}{n^n}x^n$$ is also $e$. Is it true that $\lim\limits_{n\to\infty} \sqrt[n]{a_n}$ is the radius of convergence of $\sum_{n=1}^\infty \frac{1}{a_n} x^n$ when  $a_n$ is a positive sequence of real numbers so that $\lim\limits_{n\to\infty} \sqrt[n]{a_n}$ exists and if so what would be a proof of this fact? If not, could this be modified to make it true and if so, how? What would be a proof of the modified fact?
 A: Theorem. Let $\{a_n\}_{n\geq n_0}$ a real sequence, such that $a_n>0$ at least definitely. If  $\lim_{n\to +\infty}\sqrt[n]{a_n}$ and $\lim_{n\to +\infty}\frac{a_{n+1}}{a_n}$ exist, then their limits are equal. Here tje proof by Penn: https://youtu.be/8fI0S-HeYrQ.
So, we can compute the limit:
$$\lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=\lim_{n \to +\infty}\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}=\lim_{n\to +\infty}\frac{(n+1)\cdot n!}{(n+1)!}\cdot\left(\frac{n+1}{n}\right)^n=\lim_{n\to +\infty}\left(1+\frac{1}{n}\right)^n=e$$
So:
$$\lim_{n\to +\infty}\sqrt[n]{\frac{n^n}{n!}}=e$$
A: Using Riemann Sums.
Assume the limit is $l$.
Take log on both sides
Then you have $$\ln(l)=\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}\ln(\frac{n}{r})$$.
(we are just writing the factorial as a product and then using properties of log)
Now it is easy to see that the RHS =
$\int_{0}^{1}\ln(\frac{1}{x})dx = 1$
So we have $l=e^{1}=e$
Otherwise
You have from Cauchy's First limit theorem That if $a_{n}\to a $ then $\frac{1}{n}\sum_{r=1}^{n}a_{r}\to a$ . Here is a proof of this :-Proof
This also gives that if $a_{n}>0$ and $a_{n}\to a$.
$$\ln(a)=\lim_{n\to\infty}\ln(a_{n})=\frac{1}{n}\sum_{r=1}^{n}\ln(a_{n})=\ln(\sqrt[n]{\prod_{r=1}^{n}a_{r}})$$
So $\lim_{n\to\infty}\sqrt[n]{\prod_{r=1}^{n}a_{r}}= a$.
i.e. The geometric mean and arithmetic mean tend to the same limit as that of the original sequence. These are Cauchy's first and 2nd limit theorem.
Now if $y_{1}=a_{1}$ and $y_{n}=\frac{a_{n}}{a_{n-1}}$ . Then using the 2nd limit theorem we get if $$\lim_{n\to\infty}y_{n}=\lim_{n\to\infty}\frac{a_{n}}{a_{n-1}}=l$$
Then
$$\lim_{n\to\infty}\sqrt[n]{a_{n}}=\lim_{n\to\infty}\frac{a_{n}}{a_{n-1}}=l$$.
So we want to use this.
$$\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=\lim_{n \to \infty}\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}=\lim_{n\to \infty}\frac{(n+1)^n}{n^n}=e$$.
A: With  a simpler use of Stirling's formula: we know that $n!$ is asymptotically equivalent to $\sqrt{2\pi n}\Bigl(\dfrac n{\mathrm e}\Bigr)^n$, hence
$$(n^n/n!)^{1/n}= n\,\frac 1{\bigl(n!\bigr)^{1/n}}\sim_{\infty}\not n\,\frac 1{\sqrt[2n]{2\pi n}\dfrac{\not n} {\mathrm e}}=\underbrace{\frac 1{\sqrt[2n]{2\pi n}}}_{\substack{\downarrow\\1}}\,\mathrm e.$$
A: The ratio test is quite simple if you go through logarithms
$$a_n=\left(\frac{n^n}{n!}\right)^{\frac{1}{n}}\implies \log(a_n)=\frac{1}{n}\Big[n \log(n)-\log(n!)\Big]$$ Using Stirling approximation
$$ \log(a_n)=1-\frac{\log (2 \pi  n)}{2 n}-\frac{1}{12 n^2}+O\left(\frac{1}{n^4}\right)$$ Do the same for $a_{n+1}$ and continue with Taylor series
$$\log(a_{n+1})-\log(a_n)=\frac{\log (2 \pi  n)-1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{a_{n+1} } {a_n}=e^{\log(a_{n+1})-\log(a_n) }\sim\exp\Bigg[\frac{\log (2 \pi  n)-1}{2 n^2} \Bigg]$$
