Asymptotic behavior of $\sum_{k=1}^n k^{n-k+1}$ as $n$ goes to $+\infty$

Question

I am in front of the sum $\sum_{k=1}^n k^{n-k+1}$.

I find no way to evaluate it in closed form. Is there any?

Is there a way to find an equivalent as $n$ goes to infinity?

Answer 1

Long comment

This is sequence $A003101$ in $OEIS$.

Answering this question, @leonbloy provided an extremely good approximation

$$S_n=\sum_{k=1}^n k^{n-k+1}$$ $$\log (S_n) \approx (n+2-x_0) \log(x_0) + \frac{1}{2}\log\left({\frac{2 \pi}{n+1+x_0}}\right)$$ where $$x_0=\frac{n+1}{W(e (n+1))}=\frac 1e e^{W(e (n+1))}$$ $W(.)$ being Lambert function.

Checking, if $n>7$ the maximum relative error for $S_n$ is smaller than $0.05$% and smaller than $0.02$% if $n >111$.

A few values $$\left( \begin{array}{ccc} n & \text{approximation} & S_n \\ 10 & 49845 & 49863 \\ 15 & 190691716 & 190780212 \\ 20 & 2281724622065 & 2282779990494 \\ 25 & 66042622781960110 & 66071772829247409 \\ 30 & 3938037218650692634194 & 3939676100069476203757 \\ \end{array} \right)$$

Answer 2

It is unlikely the a closed form exists, but we can calculate some asymptotics. Observe that the largest term is the $k\approx n/\log(n)$ term, which has size $\exp((1+o(1))n\log(n))$. Since there are at most $n = \exp(\log(n))$ terms, we have that the sum is asymptotically equal to $\exp((1+o(1))n\log(n))$. One could calculate more precise asymptotics, but that takes some work and you should ask yourself why you need them in the first place.