What is the value of $$\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}\frac{\sqrt[k]{k!}}{k}?$$

  • 1
    $\begingroup$ Hint: Cesaro${}{}{}$. $\endgroup$
    – Julien
    Jun 9, 2013 at 13:49

3 Answers 3


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 $\frac1e$.


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 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 $1/e$.


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