# Double summation geometric

I'm trying to find out the solution to this double summation summation:

$\sum_{x=0}^\infty \sum_{y=0}^\infty p^{x+y}$

Factoring out $p^x$ first, then solving each individual geometric series,

I get $1/(1-p)^2$. Is this correct? Or is there something wrong in my steps?

• The answer is correct! product of some of 2 infinite GPs with same common ratio should be that only! Oct 18 '17 at 17:22

## 1 Answer

Your result is correct, but you have to be careful! The sum doesn't converge for all $p$!

The sum $$\sum_{y=0}^\infty p^{x+y} = p^x\sum_{y=0}^\infty p^y=p^x\frac{1}{1-p}$$ converges if and only if $|p|<1$, and once you calculate that sum, the outer sum $$\sum_{x=0}^\infty \sum_{y=0}^\infty p^{x+y} = \frac{1}{1-p}\sum_{x=0}^\infty p^{x} = \frac{1}{(1-p)^2}$$ converges, again, for $|p|<1$.

• okay :) thank you 5xum Oct 18 '17 at 17:38