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

Question

Is there any general formula to sum following series:

$$S = 1^1 + 2^2 + 3^3 + \dotsb+(n - 1)^{n - 1} + n^n, n \in N$$

I mean for $S = f(n)$, is there a formula to compute $f(n)$?

Answer 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 $$ \frac1{n^k}\frac{(n-k)^{n-k}}{n^{n-k}} =e^{-k}\left(\frac1{n^k}+\frac{k^2}{2n^{k+1}}+\frac{3k^4+4k^3}{24n^{k+2}}+O\left(\frac1{n^{k+3}}\right)\right) $$

Answer 2

From simple graphing, I've managed to find really good bounds for your function:

$$\log_{10}(f(n))\le\frac{\ln(n^n)}{2.301}$$

I have no guarantee that my inequalities will hold true, but they most certainly hold true from graphing for $20\le n\le143$, which is decently large.

If your wondering how I derived this, I took a guess at how fast $f(n)$ grew and attempted to derive a formula with equivalent growth, seeing that $n^n<f(n)<(n+1)^{n+1}$, we can see how fast it grows, and I just took the log of it to keep numbers manageable.

From a nice article, I took the most understandable parts of it to see that for $n>2$,

$$n^n\left(\frac{4n-3}{4n-4}\right)\le f(n)\le n^n\left(\frac{2+e(n-1)}{e(n-1)}\right)$$

At proposition 2.1.