The value of the $\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}\frac{\sqrt[k]{k!}}{k}$ What is the value of  $$\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}\frac{\sqrt[k]{k!}}{k}?$$
 A: Use Stirling and Cesàro:
As $k!\approx k^ke^{-k}\sqrt{2\pi k}$, the $k$th summand is $\approx\frac1e$, hence the limit is also $\frac1e$.
A: Let $x_k=\dfrac{k!}{k^k}$
Then $$\lim_{k\to\infty}\frac{x_{k+1}}{x_k}=\lim_{k\to\infty}\frac{(k+1)!k^k}{k!(k+1)(k+1)^k}=\lim_{k\to\infty}\left(1+\frac 1k \right)^{-k}\to e^{-1}$$
This implies${}^{(1)}$ that $$\lim_{k\to\infty} x_k^{1/k}=e^{-1}$$
Then, since $x_k^{1/k}=a_k$ converges${}^{(2)}$ and has value $e^{-1}$ $$\sigma_n=\frac 1 n\sum_{k=1}^n a_k$$ also converges, and has value $$e^{-1}$$
$(1)$: Follows from $$\liminf \frac{x_{n+1}}{x_n}\leq \liminf x_n^{1/n}\leq \limsup x_n^{1/n}\leq \limsup \frac{x_{n+1}}{x_n}$$
$(2)$: Follows from $$\liminf {x_n}\leq \liminf \sigma_n \leq \limsup  \sigma_n \leq \limsup  {x_n}$$
where $\displaystyle \sigma_n:=\frac 1 n \sum_{k=1}^n x_k$.
A: A way I think about this is to break this sum up into two pieces: those for which the Stirling approximation for large $k$ is accurate, and those for which it is not.  Let such a value of $k$ dividing these pieces be $k_0$; then write
$$\lim_{n \to \infty}\frac{1}{n} \left [\sum_{k=1}^{k_0} \frac{(k!)^{1/k}}{k} + \sum_{k=k_0+1}^{n} \frac{(k!)^{1/k}}{k} \right ]$$The value of the first sum is fixed and therefore the limit is zero.  The value of the second sum, using Stirling's approximation, goes as
$$\lim_{n \to \infty}\frac{1}{n} \frac{n-k_0}{e} = \frac{1}{e}$$
The limit is therefore $1/e$.
