Sum of $p\sum_{i = 1}^{\infty} i(1 - p)^{i - 1}$ Does anyone know how to find the exact sum of

$$
p\sum_{i = 1}^{\infty} i(1 - p)^{i - 1}
$$
Note that

$$\sum_{i=1}^{\infty} i \, r^{i-1} = \frac{d}{dr} \sum_{i=0}^{\infty} r^i = \frac{d}{dr} \frac{1}{1-r}=\frac{1}{(1-r)^2}$$

So your sum is, letting $r=1-p$,

$$p \frac1{p^2} = \frac1{p}$$
Given that $|x|&lt; 1$, $$\frac 1{(1-x)^2}=\sum_{k=0}^\infty (k+1)x^k$$

So $$p^{-1}=\frac{p}{(1-(1-p))^2}=p\sum_{k=0}^\infty (k+1)(1-p)^k$$

Reasoning: $(1-x)(1+x+x^2+\ldots)=1+x+x^2+\ldots-(x+x^2+\ldots)=\lim_{n\to\infty}1-x^n=1$ so $\frac{1}{1-x}=\sum_{k=0}^\infty x^k$.  Note that $$\frac{1}{(1-x)^2}=\sum_{k=0}^\infty \frac{x^k}{1-x}=\sum_{k=0}^\infty\left(\sum_{\ell=0}^\infty x^{k+\ell}\right)$$
For any $c$, there are $c+1$ pairs $(k,\ell)$ such that $k+\ell=c$, so the sum reduces to $\sum_{k=0}^\infty (k+1)x^k$