Double summation with a variable stopping point

Question

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.

Answer 1

This problem involves sums of geometric sequences for which there is a well-known formula:

$$\sum_{n=0}^{m-1} r^n = \frac{1-r^m}{1-r}.$$

Applying this to your summation and using a little algebra you get:

$$\begin{equation} \begin{aligned} \sum_{n=2}^\infty \sum_{k=0}^{n-2} \Big( \frac{1}{4} \Big)^k \Big( \frac{1}{2} \Big)^{n-k-2} &= \sum_{n=2}^\infty \Big( \frac{1}{2} \Big)^{n-2} \sum_{k=0}^{n-2} \Big( \frac{1}{4} \Big)^k \Big( \frac{1}{2} \Big)^{-k} \\[6pt] &= \sum_{n=2}^\infty \Big( \frac{1}{2} \Big)^{n-2} \sum_{k=0}^{n-2} \Big( \frac{1}{2} \Big)^k \\[6pt] &= \sum_{n=2}^\infty \Big( \frac{1}{2} \Big)^{n-2} \cdot \frac{1-(\tfrac{1}{2})^{n-1}}{1-\tfrac{1}{2}} \\[6pt] &= 2 \Bigg[ \sum_{n=2}^\infty \Big( \frac{1}{2} \Big)^{n-2} - \sum_{n=2}^\infty \Big( \frac{1}{2} \Big)^{2n-3} \Bigg] \\[6pt] &= 2 \Bigg[ \sum_{n=0}^\infty \Big( \frac{1}{2} \Big)^{n} - \frac{1}{2} \sum_{n=0}^\infty \Big( \frac{1}{4} \Big)^{n} \Bigg] \\[6pt] &= 2 \Bigg[ \frac{1}{1-\tfrac{1}{2}} - \frac{1}{2} \cdot \frac{1}{1-\tfrac{1}{4}} \Bigg] \\[6pt] &= 2 \Bigg[ 2 - \frac{1}{2} \cdot \frac{4}{3} \Bigg] \\[6pt] &= 2 \Bigg[ 2 - \frac{2}{3} \Bigg] \\[6pt] &= 2 \cdot \frac{4}{3} \\[6pt] &= \frac{8}{3} = 2.6\bar{6}. \\[6pt] \end{aligned} \end{equation}$$

Answer 2

$$ \begin{align} \sum_{n=2}^\infty\sum_{k=0}^{n-2}\left(\frac14\right)^k\left(\frac12\right)^{n-k-2} &=\sum_{n=0}^\infty\sum_{k=0}^n\left(\frac14\right)^k\left(\frac12\right)^{n-k}\tag1\\ &=\sum_{k=0}^\infty\sum_{n=k}^\infty\left(\frac14\right)^k\left(\frac12\right)^{n-k}\tag2\\ &=\sum_{k=0}^\infty\sum_{n=0}^\infty\left(\frac14\right)^k\left(\frac12\right)^n\tag3\\ \end{align} $$ Explanation:
$(1)$: substitute $n\mapsto n+2$
$(2)$: switch order of summation ($n\ge k$)
$(3)$: substitute $n\mapsto n+k$

Does this look easier?

Answer 3

\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}