Any simplification of $\sum_{k=0}^{n}\binom{n}{k} \frac{(-1)^{k}}{(N-k)^2}$

Question

I am trying to find a compact expression for $\sum_{k=0}^{n}\binom{n}{k} \frac{(-1)^{k}}{(N-k)^2}$, where $N > n \geq 0$. Maple simplifies it using hypergeometric functions, which is not very useful. Any idea whether this can be simplified ? and if yes, how ? Thanks a lot!

Answer 1

In order to evaluate where $N\gt n$

$$\sum_{k=0}^n \frac{(-1)^k}{(N-k)^2} {n\choose k}$$

we introduce the function

$$f(z) = \frac{n! (-1)^n}{(N-z)^2} \prod_{q=0}^n \frac{1}{z-q}.$$

This has the property that with $0\le k\le n$

$$\;\underset{z=k}{\mathrm{res}}\; f(z) = \frac{n! (-1)^n}{(N-k)^2} \prod_{q=0}^{k-1} \frac{1}{k-q} \prod_{q=k+1}^n \frac{1}{k-q} \\ = \frac{n! (-1)^n}{(N-k)^2} \frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!} \\ = \frac{(-1)^k}{(N-k)^2} {n\choose k}.$$

Now residues sum to zero and the residue at infinity is zero so we may evaluate using minus the residue at $z=N$. We find

$$- n! (-1)^n \left.\left[ \prod_{q=0}^n \frac{1}{z-q} \right]'\right|_{z=N} \\ = n! (-1)^n \left. \prod_{q=0}^n \frac{1}{z-q} \sum_{q=0}^n \frac{1}{z-q} \right|_{z=N} \\ = n! (-1)^n \prod_{q=0}^n \frac{1}{N-q} \sum_{q=0}^n \frac{1}{N-q} \\ = n! (-1)^n \frac{(N-n-1)!}{N!} (H_N - H_{N-n-1}).$$

This is

$$\bbox[5px,border:2px solid #00A000]{ \frac{(-1)^n}{n+1} {N\choose n+1}^{-1} (H_N - H_{N-n-1}).}$$

Here we can nicely observe the poles when $N\le n$ (inverse binomial coefficient).