Using the Cauchy product to derive a closed form for a binomial series

Question

How can one prove hat $\forall N\in \mathbb{N}$, and $z\in \mathbb{C}$ with $|z|<1,$ $$\frac{1}{(1-z)^{N+1}}=\sum\limits_{n=0}^\infty \dbinom{N+n}{n}z^n$$ by using Cauchy product? Should I somehow split the right hand side in two parts?

Answer 1

Hint: Use the formula for a geometric series, $\frac{1}{1 - z} = \sum_{n \geq 0} z^n$, and induction. For instance, to prove the result for $N= 1$, we have $$ \frac{1}{(1 -z)^2} = \frac{1}{1 - z} \frac{1}{1 - z} = \left(\sum_{n \geq 0} z^n \right)\left(\sum_{n \geq 0} z^n \right) \, . $$ The formula for the Cauchy product of two series states that the coefficient of $z^n$ in the product $\left(\sum_{n \geq 0} a_n z^n \right) \left(\sum_{n \geq 0} b_n z^n \right)$ is $\sum_{k = 0}^n a_{n-k}b_k$. All the coefficients for the above series are $1$, so the $z^n$ coefficient is $\sum_{k=0}^n 1 \cdot 1 = n + 1 = \binom{1 + n}{n}$, as desired.

Can you do the inductive step? You may need to use the identity $\sum_{k=0}^n \binom{N+k}{k} = \binom{N+n+1}{n}$.