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Some time ago in lecture I saw the claim that if we let $p = m + n$ then

$$\sum_{n=0}^\infty \sum_{m=0}^\infty [\cdots] = \sum_{p=0}^\infty \sum_{n=0}^p [\cdots]$$

However I'm struggling to see why this is, could anyone give me a brief explanation?

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2 Answers

up vote 2 down vote accepted

This depends on the content of $[..]$. Usually the context in power series, i.e. $\sum a_nx^n \sum b_mx^m=\sum_p(\sum_{n=0}^pa_nb_{p-n})x^p$.

To see why it is so, think what happens when the series are finite but sum up to a large $N$, and what happens for $p$ much smaller than $N$. You'll see that the coefficient of $x^p$ is determined only by a finite number of coefficients; namely, those of $x^t$ for all $t\le p$, and it is determined exactly by the equation $\sum_{n=0}^pa_nb_{p-n}$.

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This as the convolution operation on ordinary generating functions as well :) –  saadtaame Feb 26 '13 at 12:52
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They just changed the order of summation. Draw a lattice of the first quadrant : i.e. draw $$ \{ (m,n) \, | \, m,n\ge 0 \}. $$ On the LHS, the sum over $m$ is a sum over an horizontal line, and then the sum over $n$ sums over all those lines. On the RHS, the sum that goes from $n = 0$ to $p$ is a sum over the diagonals $D_p = \{ (m,n) \, | \, m+n = p \}$ so that given $n$, you have $m = p-n$. (The function $f(x) = p-x$ is a diagonal that intersects the $x$ and $y$ axis at positive values.) Letting $n$ from $0$ to $p$ means $m$ goes from $p$ to $0$ (in the other direction) and summing over $p$ is just summing over all diagonals.

There is still the problem of convergence, but I can't look at this since the arguments you took were $[\dots]$ and I don't know what that is. =P

Hope that helps,

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