Is there any general formula for $S = 1^1 + 2^2 + 3^3 + \dotsb+(n - 1)^{n - 1} + n^n, n \in N$? [duplicate]

Solution 1:

I don't know if a closed form exists, but an asymptotic approximation can be given as $$ \begin{align} &n^n\left(1+\frac1n\frac{(n-1)^{n-1}}{n^{n-1}}+\frac1{n^2}\frac{(n-2)^{n-2}}{n^{n-2}}+\frac1{n^3}\frac{(n-3)^{n-3}}{n^{n-3}}+O\left(\frac1{n^4}\right)\right)\\ &=\bbox[5px,border:2px solid #C0A000]{n^n\left(1+\frac1{en}+\frac{3+e}{2e^2n^2}+\frac{52+60e+7e^2}{24e^3n^3}+O\left(\frac1{n^4}\right)\right)} \end{align} $$ where $$ \begin{align} \log\left(\frac{(n-k)^{n-k}}{n^{n-k}}\right) &=(n-k)\left(-\frac kn-\frac{k^2}{2n^2}-\frac{k^3}{3n^3}+O\left(\frac1{n^4}\right)\right)\\ &=-k+\frac{k^2}{2n}+\frac{k^3}{6n^2}+O\left(\frac1{n^3}\right) \end{align} $$ and so $$ \frac1{n^k}\frac{(n-k)^{n-k}}{n^{n-k}} =e^{-k}\left(\frac1{n^k}+\frac{k^2}{2n^{k+1}}+\frac{3k^4+4k^3}{24n^{k+2}}+O\left(\frac1{n^{k+3}}\right)\right) $$