I am trying to derive the formula for the variance of a geometric distribution and am stuck at the following problem:
I need to find the sum to infinity for the following series:
$1+3(1-p)+5(1-p)^2+7(1-p)^3+\dots$ where $p$ where $p$ is a constant $0<p<1$
The series looks like the sum of the odd numbers and the sum of a geometric series (with common ration $(1-p)$) combined.
I don't know how to deal with such series. Please tell me how to solve this.
Best Answer
Write $q = \sqrt{1-p}$. Then your series becomes
$$1\cdot q^0 + 3\cdot q^2 + 5\cdot q^4 + 7\cdot q^6 +\dotsb = \sum_{k=0}^\infty (2k+1)\cdot q^{2k}.$$
The latter is easily recognised as
$$\frac{d}{dq} \sum_{k=0}^\infty q^{2k+1} = \frac{d}{dq} q\sum_{k=0}^\infty (q^2)^k = \frac{d}{dq} \frac{q}{1-q^2} = \frac{(1-q^2) -(-2q)q}{(1-q^2)^2} = \frac{1+q^2}{(1-q^2)^2}.$$
Now insert $q^2 = 1-p$ to obtain
$$\frac{2-p}{p^2}.$$