An identity involving binomial coefficients and rational functions

Question

While going through some exercises in my analysis textbook, I came up with an equation which looks like an identity. I strongly believe that this is the case, but I couldn't prove this.

$$\sum_{0\leq k\leq n}(-1)^k\frac{p}{k+p}\binom{n}{k} = \binom{n+p}{p}^{-1}$$

Can someone provide a proof of this identity? Also, it would help a lot if you could explain the general strategy of proving such identities, if there is one, for me. Thank you.

Answer 1

The general strategy is to introduce a variable $x$ and then to deal with some polynomials and/or other power series, of which your expression is a value at $x=1$.

In you case it goes along these lines: $$\sum_{0\leq k\leq n}(-1)^k\frac{p}{k+p}\binom{n}{k}=F(1),\text{ where} \\ F(x)=\sum_{0\leq k\leq n}(-1)^kx^{k+p}\frac{p}{k+p}\binom{n}{k}$$

The power was chosen out of the blue, so that the function would simplify a little when we take a derivative. $$F'(x)=p\cdot\sum_{0\leq k\leq n}(-1)^kx^{k+p-1}\binom{n}{k}=p\cdot x^{p-1}\sum_{0\leq k\leq n}(-x)^k\binom{n}{k}=p\,x^{p-1}(1-x)^n$$

Together with the obvious $F(0)=0$ and the knowledge of beta function, this gives us the way to reconstruct the desired $F(1)$: $$F(1)=\int\limits_0^1F'(x)dx=p\int\limits_0^1x^{p-1}(1-x)^ndx=p\cdot B(p,n+1)=\\ =p\cdot{(p-1)!\;n!\over(n+p)!}={p!\;n!\over(n+p)!}=\binom{n+p}{p}^{-1}$$

Answer 2

This identity has appeared on MSE on several occasions in various forms. There is a proof using residues which goes like this (quoted from what should be earlier posts). Start with the function

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

We then get

$$\mathrm{Res}_{z=k} f(z) = n! (-1)^n \frac{p}{k+p} \prod_{q=0}^{k-1} \frac{1}{k-q} \prod_{q=k+1}^n \frac{1}{k-q} \\ = n! (-1)^n \frac{p}{k+p} \frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!} = (-1)^k \frac{p}{k+p} {n\choose k}.$$

Residues sum to zero and hence we have

$$\sum_{k=0}^n (-1)^k \frac{p}{k+p} {n\choose k} = - \mathrm{Res}_{z=\infty} f(z) - \mathrm{Res}_{z=-p} f(z).$$

Observe that $\lim_{R\to\infty} 2\pi R/R/R^{n+1} = 0$ and the residue at infinity is zero. It remains to compute

$$- \mathrm{Res}_{z=-p} f(z) = - n! (-1)^n \times p \times \prod_{q=0}^n \frac{1}{-p-q} \\ = n! \times p \times \prod_{q=0}^n \frac{1}{p+q} = n! \times p \times \frac{(p-1)!}{(p+n)!} = {n+p\choose p}^{-1}.$$

This concludes the argument. Here we require that $-p$ not be among the poles in $[0,n],$ resulting in a singularity in the sum.