How can I solve a sum of permutations mutiplied by the argument?

Question

I have to solve this:

$\sum_{k=0}^{n-1} \frac{k\cdot n!}{(n-k)!}$

I am trying to decompose it somehow, but so far, no luck.

Answer 1

$S=\sum\limits_{k=0}^{n-1} \dfrac{k\cdot n!}{(n-k)!}=\sum\limits_{k=0}^{n-1}\binom{n}{k}k!k\tag1$

Known by the definition of $\Gamma$ that:

$k!=\Gamma(k+1)= \int\limits_0^{\infty}x^k e^{-x}dx$ put it back to (1) we have:

$\sum\limits_{k=0}^{n-1}\binom{n}{k}\int\limits_0^{\infty}kx^k e^{-x}dx=\int\limits_0^{\infty}xe^{-x}\frac{d}{dx}\sum\limits_{k=0}^{n-1}\binom{n}{k}x^k dx\tag2$

Replaced the orders and using that $kx^{k-1}=\frac{dx^k}{dx}$.

Based on Binomial theorem we can have: $\sum\limits_{k=0}^{n-1}\binom{n}{k}x^k=(x+1)^n-x^n \tag3$

We get the following integral:

$\int\limits_0^{\infty}xe^{-x}n\Big[(x+1)^{n-1}-x^{n-1}\Big] dx=\int\limits_0^{\infty}xe^{-x}n(x+1)^{n-1}dx-\int\limits_0^{\infty}e^{-x}n x^{n}dx$

Using the definition of $\Gamma(z)$ and upper incomplete $\Gamma(z,s)$ we get the result:

$S=n(e \Gamma(n+1,1)-e \Gamma(n,1)-\Gamma(n))\tag4$

where further simplification is possible.