Can I get better approximation of $\sum_{k=1}^{n} k^k$ Is it possible to get approximation$f(n)$ of $\sum_{k=1}^{n} k^k$ with
\begin{align}
\lim_{n\to +\infty }\left(f(n)-\sum_{k=1}^{n} k^k\right)=0
\end{align}
Thanks for your attention!
 A: I have no idea how $f(n)$ should look like, and actually I suspect that there is no such simple formula for $f(n)$. But at least we can improve Sasha's asymptotics. Indeed, a moment of thought gives that for any fixed $m$, we have
\begin{align*}
\frac{1}{n^n} \sum_{k=1}^{n} k^k
&= \sum_{k=0}^{n-1} \left(1 - \frac{k}{n}\right)^{n-k}\frac{1}{n^k} \\
&= \sum_{k=0}^{n-1} \frac{1}{(en)^k}\exp\bigg\{k \sum_{j=1}^{\infty} \frac{(k/n)^j}{j(j+1)} \bigg\} \\
&= \sum_{k=0}^{m} \frac{1}{(en)^k}\exp\bigg\{k \sum_{j=1}^{m-k} \frac{(k/n)^j}{j(j+1)} \bigg\} + \mathcal{O}(n^{-(m+1)})
\end{align*}
Plugging $m = 1$, we have
$$ \frac{1}{n^n} \sum_{k=1}^{n} k^k = 1 + \frac{1}{en} + \mathcal{O}(n^{-2}) $$
With aid of Mathematica, for $m = 4$ we have
\begin{align*}
\frac{1}{n^n} \sum_{k=1}^{n} k^k
&= 1 + \frac{1}{en} + \left( \frac{1}{e^2}+\frac{1}{2e}\right) \frac{1}{n^2} + \left(\frac{1}{e^3}+\frac{2}{e^2}+\frac{7}{24 e}\right)\frac{1}{n^3} \\
&\qquad + \left(\frac{1}{e^4}+\frac{9}{2e^3}+\frac{10}{3 e^2}+\frac{3}{16 e}\right)\frac{1}{n^4} + \mathcal{O}(n^{-5})
\end{align*}
