<s>[INST] Find constant that satisfies joint probability mass function for $X$ and $Y$ 
$f(x,y) = c(1/4)^x(1/3)^y$, $x = 1,.... y= 1,....$
The solution is $6$, but I am having trouble coming to this, and I think my issue has to do with $x$ and $y$ being infinitely countable. I would be grateful for any explanations.
 [/INST] $$\sum_{x=1}^\infty \sum_{y=1}^\infty c\left(\frac{1}{4}\right)^x\left(\frac{1}{3}\right)^y 
= c\sum_{x=1}^\infty \left(\frac{1}{4}\right)^x \times \sum_{y=1}^\infty \left(\frac{1}{3}\right)^y = c\left(\frac{1}{4-1}\right)\left(\frac{1}{3-1}\right) = c\left(\frac{1}{3}\right)\left(\frac{1}{2}\right) = c\left(\frac{1}{6}\right)= 1.$$ And, $$ c = 6 $$
Thanks, my issue was getting formula for geometric series:  $$ \sum_{x=1}^\infty \left(\frac{1}{x^n}\right)= \left(\frac{1}{x-1}\right) $$
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