Double summation with a variable stopping point I am interested in calculating the following double summation:
$\sum_{n=2}^ \infty \sum_{k =0}^{n-2}\frac{1}{4}^k \frac{1}{2}^{n-k-2}$
I don't really know where to start, so I was hoping someone could point me to some resource where I could learn the terminology/methodology associated with solving such problems.
The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context previously.

\begin{aligned}
\sum_{n=2}^ \infty \sum_{k =0}^{n-2}\left(\frac{1}{4}\right)^k \left(\frac{1}{2}\right)^{n-k-2} &= \sum_{n=2}^ \infty \sum_{k =0}^{n-2}\left(\frac{1}{2}\right)^{2k} \left(\frac{1}{2}\right)^{n-k-2} \\
&= \sum_{n=2}^ \infty \sum_{k =0}^{n-2}\left(\frac{1}{2}\right)^{n+k-2} \\
&= \sum_{n=2}^ \infty \left(\frac{1}{2}\right)^{n-2}\sum_{k =0}^{n-2}\left(\frac{1}{2}\right)^{k}  \\
&= \sum_{n=2}^ \infty \left(\frac{1}{2}\right)^{n-2} \frac{1-\left(\frac 12\right)^{n-1}}{1-\frac 12} \\
&= \sum_{n=2}^ \infty\left(\frac{1}{2}\right)^{n-3} - \left(\frac{1}{2}\right)^{2n-4} \\
&= 2 \sum_{n=0}^ \infty\left(\frac{1}{2}\right)^n -\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^n \\
&= 2 \frac{1}{1-\frac 12}-\frac{1}{1-\frac 1 4} \\
&= \frac 8 3
\end{aligned}
