How to show that two expressions containing 2-dimensional sums with different iterators are equal

Question

I am trying to prove that these two expressions are equal (I verified numerically) but it's not obvious to me at first sight why they are (might be to some).

$$\sum_{k=0}^{\infty}\sum_{i=0}^k{f(i,k-i)}=\sum_{j=0}^\infty\sum_{i=0}^\infty{f(i,j)}$$

It's a simple substitution of one iterator with another ($j=k-i$) but I'm not sure how to make that substitution when they are iterators.

Could someone please explain why this is evident or provide a method of elaboration of some sort that shows it?

Answer 1

Fixing $k=k_0$ at the left hand side we have as inner sum \begin{align*} \sum_{i=0}^{k_0}f(i,k_0-i)=f(0,k_0)+f(1,k_0-1)+\cdots+f(k_0,0) \end{align*} which is a section of a minor diagonal. On the other hand the right hand side gives for $j=j_0$ \begin{align*} \sum_{i=0}^\infty f(i,j_0)=f(0,j_0)+f(1,j_0)+f(2,j_0)+\cdots \end{align*} the vertical half-line starting at $(0,j_0)$.

Assuming the double series is unconditionally convergent we can rearrange the terms and obtain \begin{align*} \sum_{j=0}^\infty\sum_{i=0}^\infty f(i,j)&=\sum_{k=0}^\infty \left(\sum_{{i+j=k}\atop{i,j\geq 0}}f(i,j)\right)\tag{1}\\ &=\sum_{k=0}^\infty\sum_{i=0}^kf(i,k-i)\tag{2} \end{align*} and the claim follows.

Comment:

• In (1) we introduce an index variable $k\geq 0$ and collect terms $f(i,j)$ with $i+j=k$.

• In (2) we get rid of the index $j$ by using $i+j=k$.