# Why does this infinite sum compute the generalized binomial coefficient?

Suppose we define the binomial coefficient for two complex values as

$$\binom{x}{z} = \frac{\Gamma(x+1)}{\Gamma(z+1) \cdot \Gamma(x-z+1)}$$

where $\Gamma(x)$ is the gamma function.

I no longer recall what specific process led me to the following identity, but a while ago, working with Mathematica, a bunch of symbolic manipulation led to the following

$$\binom{x}{z} = \frac{\sin(\pi z)}{\pi} \cdot \sum_{k=0}^{\infty} \frac{(-1)^k}{z-k} \cdot \binom{x}{k}$$

which a cursory attempt at empirical evidence seems to back up.

Can anyone explain to me why the term on the right here equals the generalized binomial coefficient, if it really does?

• Because the gamma function generalises the factorial, i.e. $\Gamma(n)=(n-1)!$ for positive integers. – Levent Nov 19 '17 at 22:59

Let us start with the simple part:

$$\frac{\sin \pi z}{\pi}=\frac{1}{\Gamma(z) \Gamma(1-z)}$$

Now we move on to the series:

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{z-k} \cdot \binom{x}{k}$$

Consider the ratio of successive general terms:

$$\frac{T_{k+1}}{T_k}=\frac{(-1)(z-k)(x-k)}{(z-k-1)(k+1)}=\frac{(k-x)(k-z)}{(k+1-z)} \frac{1}{(k+1)}$$

$$T_0=\frac{1}{z}$$

Which means:

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{z-k} \cdot \binom{x}{k}= \frac{1}{z} {_2 F_1 } (-x,-z;1-z;1)$$

Let us use the Euler integral for the Hypergeometric function:

$${_2 F_1 } (-x,-z;1-z;1)=\frac{1}{B(-z,1)} \int_0^1 t^{-z-1} (1-t)^0 (1-t)^x dt$$

In other words:

$${_2 F_1 } (-x,-z;1-z;1)=-z B(-z,1+x)$$

Finally, we can write:

$$\frac{\sin(\pi z)}{\pi} \cdot \sum_{k=0}^{\infty} \frac{(-1)^k}{z-k} \cdot \binom{x}{k}=\frac{-B(-z,1+x)}{\Gamma(z) \Gamma(1-z)}=\frac{-\Gamma(-z) \Gamma(1+x)}{\Gamma(z) \Gamma(1-z) \Gamma(1+x-z)}$$

Finally:

$$\frac{\sin(\pi z)}{\pi} \cdot \sum_{k=0}^{\infty} \frac{(-1)^k}{z-k} \cdot \binom{x}{k}=\frac{\Gamma(x+1)}{\Gamma(z+1) \Gamma(x-z+1)} \equiv \binom{x}{z}$$