Is there any kind of formula to estimate this $1^1+2^2+3^3+...+n^n$?

Question

Is there a efficient formula to calculate this sum ? $$S = 1^1+2^2+3^3+...+n^n$$ , or it simply diverges ?

Answer 1

The function $S_n=\sum_{k=1}^n k^k$ has been studied often, see here for references. The question for a formula or an estimate is the well known Problem 4155 by G. W. Wishard, published in Amer. Math. Monthly, 53 (1946), 471. A solution was given there by F. Underwood.

Edit: The linked paper shows the estimate $$ n^n\left( 1+\frac{1}{4(n-1)}\right)<S_n<n^n\left( 1+\frac{2}{e(n-1)}\right). $$ I am thankful to Ross Millikan, who has provided this estimate.

Answer 2

Claim $$\lim_{n\rightarrow \infty} \frac{1^1+2^2+3^3+\cdots+n^n}{(n+1)^n}= \frac{1}{e}$$

We will do that by using Cesaro-Stolz the discrete version of the L'Hospital's rule.

\begin{align*} \lim_{n\rightarrow \infty} \frac{1^1+2^2+3^3+\cdots+n^n}{(n+1)^n}&=^{\text{CS}}\lim_{n\rightarrow \infty} \frac{(n+1)^{n+1}}{(n+2)^{n+1} -(n+1)^n}= \\ &=\lim_{n\rightarrow \infty} \frac{1}{(1+\frac{1}{n+1})^{n+1} -(n+1)^{-1}}\\ &=\frac{1}{e} \end{align*}