How to prove that $\sum_{k=0}^{\infty}\binom{k+r}{r}p^r(1-p)^k=\frac 1p$?

Question

I want to prove $$\sum_{k=0}^{\infty}\binom{k+r}{r}p^r(1-p)^k=\frac 1p$$ for $p\in (0,1)$ and $q=1-p$, Mathematica told me it's an identity, but I have no idea to deal with the binomial coefficient, is there any way to find the sum of this series?

Answer 1

The first step is to show that $$ \frac{1}{(1 - x)^{r + 1}} = \sum_{k = 0}^\infty \binom{k + r}{r} x^k, \text{ for } |x| < 1. $$ This follows from the generalized binomial theorem: $$ (1 + x)^\alpha = \sum_{k = 0}^\infty \binom{\alpha}{k}x^k, \text{ where } \binom{\alpha}{k} = \frac{\alpha(\alpha-1) \cdots (\alpha-k+1)}{k!}. $$

What you need to show is that, where $\alpha = -(r + 1)$, $$ \binom{-(r + 1)}{k} = \binom{k + r}{k}(-1)^k.$$ Also note that $\binom{k+r}{k} = \binom{k+r}{r}$.

After all that, $$ \sum_{k = 0}^\infty \binom{k + r}{r} p^r (1-p)^k = \frac{p^r}{(1-(1-p))^{r+1}} = \frac1p. $$

Answer 2

You can multiply both sides by $p$ and then note that the summand is the probability mass function for a negative binomial distribution with success probability $p$, so the sum must be 1.