I would like to prove:

$$\left\{ \frac{1}{n^n}\sum_{k=1}^{n}{k^k} \right\}_{n \in \mathbb{N}}\rightarrow 1$$

I found a proof applying Stolz criterion but I need to use the fact that:

$$\left\{\left(\frac{n}{n+1}\right)^n\right\}\text{ is bounded}\tag{$\ast$}$$

I would like to calculate this limit without using $(\ast)$ and preferably using basic limit criterion and properties.

(Sorry about mispellings or mistakes, English is not my native language.)


Obviously $$\frac{1}{n^n}\sum_{k=1}^{n}k^k \geq 1,$$ while: $$\frac{1}{n^n}\sum_{k=1}^{n}k^k\leq \frac{1}{n^n}\sum_{k=1}^{n}n^k\leq\frac{1}{n^n}\cdot\frac{n^n}{1-\frac{1}{n}}=\frac{n}{n-1}.$$

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  • 1
    $\begingroup$ Very simple proof, thanks! $\endgroup$ – Psi Jan 9 '15 at 21:00
  • $\begingroup$ (Morally) shouldn't there be an equality sign in the second step? $\endgroup$ – Pedro Tamaroff Jan 9 '15 at 21:12
  • $\begingroup$ @PedroTamaroff: yes, sure. Now fixed. $\endgroup$ – Jack D'Aurizio Jan 9 '15 at 21:16


$$n^n\le\sum_{k=1}^nk^k\le n^n+(n-1)^{n-1}+(n-2)(n-2)^{n-2}$$

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