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I'm trying to reduce the following nested summation (removing all summations), using the fact that: $\sum_{i=0}^\infty\ a^i = 1/(1- a)$.

Problem: $\sum_{n=0}^\infty\sum_{m=n}^\infty\ a^nb^m$

I know that if instead of m=n in the summation, we had m=0, it would be easy to reduce. I'd just use the above fact twice. How could I reduce this problem, removing all summations, using the above fact?

Note: Although this IS related to a homework assignment, you are welcome to post a final answer. This is just some algebra that's a part of a much larger probability question.

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2 Answers 2

up vote 2 down vote accepted

If you don't know the general formula for the sum of a geometric series that doesn't start at $0$, you can do this

$$\sum_{m=n}^\infty b^m=\sum_{k=0}^\infty b^{n+k}=b^n\sum_{k=0}^\infty b^k=\frac{b^n}{1-b}$$

to derive it.

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Hint: $$\sum_{m=n}^\infty a^nb^m=a^n\sum_{m=n}^\infty b^m = a^n\sum_{m=n}^\infty (b^n)(b^{m-n})=a^nb^n\sum_{m=n}^\infty b^{m-n}=a^nb^n\sum_{k=0}^\infty b^k.$$

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