I have this sum which looks somewhat like a sum of a binomial distribution but not quite, and I'm not quite sure how I'd prove this equation:
$$\sum_{n = 0}^{\infty}\left(\frac{n}{mn + 1} {{mn + 1}\choose n} p^n (1-p)^{mn + 1 - n}\right) = \frac{p}{1 - mp}$$
In this equation we also know that $p = 0.0913$ and that $mp < 1 \implies m < \frac{1}{0.0913}$. Any tips or references would be appreciated, don't really know where I would start. Without the fraction in front it would be a binomial distribution sum, but I'm not sure how the fraction affects the sum
Here's a start: \begin{align} \sum_{n=0}^\infty \frac{n}{mn + 1} \binom{mn + 1}{n} p^n (1-p)^{mn + 1 - n} &=\sum_{n=1}^\infty \frac{n}{mn + 1} \binom{mn + 1}{n} p^n (1-p)^{mn + 1 - n} \\ &= \sum_{n=1}^\infty \frac{n}{mn + 1} \frac{mn + 1}{n}\binom{mn + 1-1}{n-1} p^n (1-p)^{mn + 1 - n} \\ &= \sum_{n=1}^\infty \binom{mn}{n-1} p^n (1-p)^{mn + 1 - n} \\ &= \sum_{n=0}^\infty \binom{m(n+1)}{(n+1)-1} p^{n+1} (1-p)^{m(n+1)+1 - (n+1)} \\ &= \sum_{n=0}^\infty \binom{m(n+1)}{n} p^{n+1} (1-p)^{m(n+1)-n} \\ &= p \sum_{n=0}^\infty \binom{m(n+1)}{n} p^n (1-p)^{m(n+1)-n} \end{align}