# Double summation with a variable stopping point [closed]

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.

## closed as off-topic by Namaste, Xander Henderson, Isaac Browne, Leucippus, TaroccoesbroccoJul 31 '18 at 8:29

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• Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. – robjohn Jul 31 '18 at 0:22
• 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. I will make sure to do so in the future. – Filip Jul 31 '18 at 2:13
• I have added the comment as context to your question. Thank you for doing this in the future. – robjohn Jul 31 '18 at 3:14

## 3 Answers

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}

\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?

\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}