Generating function with binomial coefficients

Question

I want to derive formula for generating function $$\sum_{n=0}^{+\infty}{m+n\choose m}z^n$$ because it is very often very useful for me. Unfortunately I'm stuck:

$$ f(z)=\sum_{n\ge 0}{m+n\choose n}z^n= \\ \sum_{n\ge 1}{m+n-1\choose n-1}z^n+\sum_{n\ge 0}{m+n-1\choose n}z^n= \\ zf(z)+\sum_{n\ge 0}{m+n-1\choose n}z^n$$

here, I'm afraid that there is a need for something more sophisticated than above trivial identity. Can you help me?

Answer 1

The function $\left(1 + z\right)^k$ has series expansion $$\left(1 + z\right)^k = \sum_{n\ge0}\binom{k}{n}z^n$$ We also have the binomial coefficient identity $$\binom{n}{k} = (-1)^k\binom{-n + k - 1}{k}$$ which together gives $$\sum_{n\ge 0}\binom{m+n}{n}z^n = \sum_{n\ge 0}\binom{-m-1}{n}(-z)^n = \frac{1}{(1-z)^{m+1}}$$ which is your required generating function.

Answer 2

$$f_m(z).(1-z)=\sum\left(\binom{m+n}{n}-\binom{m+n-1}{n-1}\right)z^n=\sum\binom{(m-1)+n}{n}z^n=f_{m-1}(z)$$

And obviously $f_0(z)=\frac{1}{1-z}$, so $f_m(z)=\frac{1}{(1-z)^{m+1}}$