I need to expand $$\frac{n!}{(n-k)!}$$ as a polynomial in n. I know the term with n^1 has coefficient (-1)^{k+1} the term with n^k has coeff 1, and the term with n^{k-1} has coefficient $$\frac{k(k-1)}{2}$$. Can we write this as a series for general k?

Question

I need to expand $$\frac{n!}{(n-k)!}$$ as a polynomial in n.  I know the term with n^1 has coefficient (-1)^{k+1} the term with n^k has coeff 1, and the term with n^{k-1} has coefficient $$\frac{k(k-1)}{2}$$.  Can we write this as a series for general k?

anonymous · Answer

sorry I meant to say term with n^1 has coeff (-1)^{k+1}*(k-1)!

watchmath · Answer

So basically you want to expand $(n-k+1)\cdots n$ ?

anonymous · Answer

yes the coefficient of the kth derivative of x^(n)

anonymous · Answer

im trying to solve a bigger problem and this is a part of it

watchmath · Answer

and what is the bigger problem :)

anonymous · Answer

$$\sum_{n=0}^{\infty}\frac{n^k}{e^{an}}$$

for general k>=1 and positive a.  I solved a specific case of this where k=2 a=2 for someone and am trying to come up with a general formula.

anonymous · Answer

maybe there is a simpler way than what I am doing but I'm trying to look at derivatives of geometric series.

watchmath · Answer

You mean you want to evaluate the series and not only finding out the convergence of the series?

anonymous · Answer

yes the exact sum

anonymous · Answer

i evaluated it for n=2,k=2 by hand and had maple solve for various other values and it looks like there will be a nice general solution

anonymous · Answer

sorry k=2, a=2

anonymous · Answer

I geuss I've just never seen the expansion written out generally.  I'm sure it has been done though.