I'm working on an Expected Value problem and I have arrived at the following infinite sum:
$$2\sum_{n=1}^{1} \left(\frac{5}{72}\right)\left(\frac{5}{9}\right)^{n-1}\left(\frac{5}{6}\right)^{1-n} + 3\sum_{n=1}^{2} \left(\frac{5}{72}\right)\left(\frac{5}{9}\right)^{n-1}\left(\frac{5}{6}\right)^{2-n} + 4\sum_{n=1}^{3} \left(\frac{5}{72}\right)\left(\frac{5}{9}\right)^{n-1}\left(\frac{5}{6}\right)^{3-n}+\cdot\cdot\cdot$$
I'm seeking a way to simplify and eventually evaluate this sum. I've looked at Brian M. Scott's Answer to something similar and I propose the following, however, it is flawed:
$$\prod_{n=2}^{\infty}n\sum_{i=1}^{n-1}\left(\frac{5}{72}\right)\left(\frac{5}{9}\right)^{i-1}\left(\frac{5}{6}\right)^{n-2}$$
I think my understanding of the nested product/sum notation is incorrect, since Wolfram Alpha does not appear to give something reasonable. Any help is appreciated!
Here is a straightforward computation of your double sum: \begin{align} \sum_{k=2}^\infty k \sum_{n=1}^{k-1} \frac{5}{72} \left(\frac{5}{9}\right)^{n-1} \left(\frac{5}{6}\right)^{k-1-n} &= \frac{1}{8} \sum_{k=2}^\infty k \left(\frac{5}{6}\right)^{k-1} \sum_{n=1}^{k-1} \left(\frac{2}{3}\right)^n \\ &= \frac{1}{8} \sum_{k=2}^\infty k \left(\frac{5}{6}\right)^{k-1} \frac{2/3-(2/3)^k}{1-2/3} \\ &= \frac{1}{4} \sum_{k=2}^\infty k \left(\frac{5}{6}\right)^{k-1} - \frac{1}{4} \sum_{k=2}^\infty k \left(\frac{5}{9}\right)^{k-1} \\ &= \frac{1}{4} \cdot \frac{(2-5/6)(5/6)}{(1-5/6)^2} - \frac{1}{4} \cdot \frac{(2-5/9)(5/9)}{(1-5/9)^2} \\ &= \frac{495}{64} \end{align}
Another approach is to interchange the order of summation: \begin{align} \sum_{k=2}^\infty k \sum_{n=1}^{k-1} \frac{5}{72} \left(\frac{5}{9}\right)^{n-1} \left(\frac{5}{6}\right)^{k-1-n} &= \frac{1}{8} \sum_{k=2}^\infty k \left(\frac{5}{6}\right)^{k-1} \sum_{n=1}^{k-1} \left(\frac{2}{3}\right)^n \\ &= \frac{1}{8} \sum_{n=1}^\infty \left(\frac{2}{3}\right)^n \sum_{k=n+1}^\infty k \left(\frac{5}{6}\right)^{k-1} \\ &= \frac{1}{8} \sum_{n=1}^\infty \left(\frac{2}{3}\right)^n 6(n+6)\left(\frac{5}{6}\right)^n \\ &= \frac{3}{4} \sum_{n=1}^\infty n\left(\frac{5}{9}\right)^n + \frac{9}{2} \sum_{n=1}^\infty \left(\frac{5}{9}\right)^n \\ &= \frac{3}{4} \cdot \frac{5/9}{(1-5/9)^2} + \frac{9}{2} \cdot \frac{5/9}{1-5/9} \\ &= \frac{495}{64} \end{align}