Proof of binomial series in the case N is a real number

Question

Could you please tell me how to prove the result below in the case that $n$ is not necessarily an integer ? (i.e. $n$ can be any real number so that $\binom{n+i}{i}$ is well defined). I just can't wrap my head up to the point of the manipulation of the combination of real numbers.

$$ { \sum_{i=0}^{\infty}x^i\binom{n+i}{i}=\frac{1}{(1-x)^{n+1}} } $$

Thank you very much!

Answer 1

We recall the definition of binomial coefficients below valid for real (even complex) $\alpha$: \begin{align*} \binom{\alpha}{n}:=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}\qquad\qquad \alpha\in\mathbb{C}, n\in\mathbb{N}_0 \end{align*} Using this definition we can show the validity of the binomial identity \begin{align*} \binom{-\alpha}{n}=\binom{\alpha+n-1}{n}(-1)^n\tag{1} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{i=0}^\infty\binom{n+i}{i}x^i} &=\sum_{i=0}^\infty\binom{-n-1}{i}(-1)^ix^i\tag{2.1}\\ &=\sum_{i=0}^\infty\binom{-(n+1)}{i}(-x)^i\tag{2.2}\\ &=(1-x)^{-(n+1)}\\ &\,\,\color{blue}{=\frac{1}{(1-x)^{n+1}}} \end{align*} according to the claim.

Comment:

• In (2.1) we apply the binomial identity (1).

• In (2.2) we use a binomial series expansion.