How to prove this identity? $\sum_{n=0}^{+\infty}{{n + \alpha - 1}\choose{n}}z^n.$

Question

In this thread, General formula for the power series of $\dfrac{1}{(1+x)^3}$

the_candyman posted an identity:

$$(1-z)^{-\alpha} = \sum_{n=0}^{+\infty}{{n + \alpha - 1}\choose{n}}z^n.$$

Notice that this binomial series has a negative exponent. I know that the general formula for binomial series is:

$$(1+x)^k=\sum_{n=0}^{+\infty}{\binom{k}{n}}x^n$$.

How do apply the second definition to prove the first identity?

Answer 1

We use the following definition of binomial identities valid for complex $\alpha$ and non-negative integers $n$: \begin{align*} \binom{\alpha}{n}=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}\tag{1} \end{align*} which can be found for instance as formula (5.1) in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.

We obtain \begin{align*} \color{blue}{\binom{n+\alpha-1}{n}}&=\frac{(n+\alpha-1)((n+\alpha-1)-1)\cdots(n+\alpha-1-(n-1))}{n!}\tag{2}\\ &=\frac{(n+\alpha-1)((n+\alpha-2)\cdots\alpha}{n!}\\ &=(-1)^n\frac{(-\alpha)(-\alpha-1)\cdots(-\alpha-(n-1))}{n!}\tag{3}\\ &\,\,\color{blue}{=(-1)^n\binom{-\alpha}{n}}\tag{4} \end{align*}

Comment:

• In (2) we use the definition (1).

• In (3) we factor out $(-1)^n$.

• In (4) we use the definition (1) again.

From (2) and (4) we conclude \begin{align*} (1-z)^{-\alpha}=\sum_{n=0}^{\infty}\binom{-\alpha}{n}(-z)^n=\sum_{n=0}^\infty\binom{n+\alpha-1}{n}z^n\qquad\qquad |z|<1 \end{align*}