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

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    Do you mean: $$f(x,y)=c\left(\frac{1}{4}\right)^x \left(\frac{1}{3}\right)^y$$ – 2017-01-21
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    Welcome to the site! I have edited your question so that your mathematical formulas are clear, and we will all see it pending moderation approval. In the future, LaTeX formatting is really quite necessary for these situations, and in general. – 2017-01-21
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    Thanks for revising the question, I will be sure to format the questions properly next time. – 2017-01-21
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    I still don't see the domain of integration. Is it $\{(x,y); x\geq 1, y\geq1\}$? – 2017-01-21
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    **x = 1, 2, . . . , y = 1, 2, . . . .** verbatim from text. So yes, I would say **{(x,y);x≥1,y≥1}** – 2017-01-21
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    @TheCount : One should not call MathJax "LaTeX". It might make people who get good at MathJax think they've mastered LaTeX. Then when they encounter actual LaTeX, they won't know what to make of things like \thispagestyle and \begin{document} and \section and \setcounter and \bibitem etc. It would be better for those who don't know LaTeX to know that they don't know LaTeX. – 2017-01-21
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    @MichaelHardy I mean... it's such a small thing... but I suppose you're right... [i guess...](http://i.imgur.com/Kp6RN00.png) – 2017-01-21
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    Nevertheless, getting proficient using MathJax can be a worthwhile step towards learning LaTeX. – 2017-01-21

2 Answers 2

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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 $c.$

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

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