Find constant that satisfies joint probability mass function for $X$ and $Y$

Question

$f(x,y) = c(1/4)^x(1/3)^y$, $x = 1,.... y= 1,....$ 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.

Answer 1

Comment: You need to have

$$\sum_{x=1}^\infty \sum_{y=1}^\infty c\left(\frac{1}{4}\right)^x\left(\frac{1}{3}\right)^y = c\sum_{y=1}^\infty \left(\frac{1}{4}\right)^x \times \sum_{y=1}^\infty \left(\frac{1}{3}\right)^y = 1.$$

Sum each geometric series separately, and then solve for $c.$

Answer 2

$$\sum_{x=1}^\infty \sum_{y=1}^\infty c\left(\frac{1}{4}\right)^x\left(\frac{1}{3}\right)^y = c\sum_{x=1}^\infty \left(\frac{1}{4}\right)^x \times \sum_{y=1}^\infty \left(\frac{1}{3}\right)^y = c\left(\frac{1}{4-1}\right)\left(\frac{1}{3-1}\right) = c\left(\frac{1}{3}\right)\left(\frac{1}{2}\right) = c\left(\frac{1}{6}\right)= 1.$$ And, $$ c = 6 $$

Thanks, my issue was getting formula for geometric series: $$ \sum_{x=1}^\infty \left(\frac{1}{x^n}\right)= \left(\frac{1}{x-1}\right) $$