Calculating infinite sums geometric series It has been a while since I've dealt with infinite sums and geometric series. Was wondering if anybody could shed some light on this and help me out. 
$ \sum_{r=1}^\infty \sum_{t=r+1}^\infty 0.02(0.9)^{r-1}(0.8)^{t-1} $
$=\sum_{r=1}^\infty 0.02(0.9)^{r-1}\frac{0.8^r}{1-0.8}$
$=\frac{0.02}{0.2}\frac{1}{0.9}\sum_{r=1}^\infty0.72^r$
$=\frac{1}{9}\frac{0.72}{1-0.72} $
$= 0.2857$  
Basically I can't seem to recall the rules for simplifying the infinite sums and need a quick refresher. Thank you for any help!
 A: Let us first look at the "inner" sum
$$\sum_{t=r+1}^\infty (0.02)(0.9)^{r-1}(0.8)^{t-1}.\tag{1}$$
The first two terms do not involve $t$, so they can be taken "outside." Our inner sum is equal to
$$(0.02)(0,9)^{r-1}\sum_{t=r+1}^\infty (0.8)^{t-1}.$$
Let us look at the sum $\sum_{t=r+1}^\infty (0.8)^{t-1}$.  The first term of this sum, when $t=r+1$, is $(0.8)^r$. The next term, when $t=r+2$, is $(0.8)^{r+1}$, and so on. So the sum $\sum_{t=r+1}^\infty (0.8)^{t-1}$ is an infinite geometric series, with first term $(0.8)^r$ and common ratio $0.8$. By I hope a familiar formula, the sum of the geometric series with first term $a$ and common ratio $t$, with $|t|\lt 1$, is equal to $\frac{a}{1-t}$. In our case $a=(0.8)^r$, and $t=0.8$, so the sum of the series $\sum_{t=r+1}^\infty 
(0.8)^{t-1}$ is $\frac{(0.8)^r}{0.2}$. It follows that the sum (1) is equal to
$$(0.02)(0.9)^{r-1}\cdot \frac{(0.8)^r}{0.2}.$$
Now we evaluate the outer sum, which is
$$\sum_{r=1}^\infty(0.02)(0.9)^{r-1}\cdot \frac{(0.8)^r}{0.2}.\tag{2}$$
I will not do it quite in the way quoted, though that is also right. Let us take a factor of $0.8$ from $(0.8)^r$, giving $(0.8)(0.8)^{r-1}$.  Our outer sum is equal to
$$\frac{0.02}{0.2}(0.8)\sum_{r=1}^\infty (0.9)^{r-1}(0.8)^{r-1}.$$
The sum $\sum_{r=1}^\infty (0.9)^{r-1}(0.8)^{r-1}$ is an infinite geometric series, first term $1$, common ratio $(0.9)(0.8)=0.72$. So this sum is $\frac{1}{1-0.72}$, and we are essentially finished. 
