## Sequences and Series, Multidimensional Series

### Multidimensional Series

Let si,j be a 2 dimensional array of elements, indexed by i ≥ 0 and j ≥ 0. One could compute the sum of s by adding along the rows, giving an infinite column of intermediate results, then add down the column to get the sum of s. This is rather like a spreadsheet. Alternatively, one could add down the columns, then add the row of intermediate results. Does it matter?

Let si,j = 1 if j = i, or -1 if j = i+1. In other words, we have 1 down the main diagonal, and -1 in the next diagonal up. Every row and every column represents an absolutely convergent series. Add along rows first and get a column of zeros, hence a sum of 0. Add along columns and get 1 for the first intermediate result, and zeros thereafter, giving a sum of 1.

Here is a better definition. Let the sum of s be the sum of the elements of s, taken in any order. Thus s is an absolutely convergent series that happenes to be displayed in 2 dimensions.

Given this definition, what happens if we add along rows, and then add those results together? We hope the answer equals the sum of s.

To simplify things,let s contain real numbers. Consider the positive terms and the negative terms in s. Let the positive terms sum to p and let the negative terms sum to q, where p ≥ 0 and q ≤ 0. We did this before, when we proved the terms of an absolutely convergent series can be rearranged without changing the sum. In fact you might want to review that proof, as this one is similar.

Every subseries converges, so let ri be the sum of si,j as j runs from 0 to infinity. Finitely many convergent series can be added, thus adding their limits. Therefore the sum of the first n terms of r is indeed the sum of all the elements in the first n rows of s, in any order you like. We need to show that rn approaches p+q.

Given an ε, add elements of s in antidiagonal stripes until the positive sum is greater than p-ε and the negative sum is less than q+ε. In other words, start with s0,0, then add s0,1+s1,0, then add s0,2+s1,1+s2,0, and so on until we are close to p and q. Assume the last antidiagonal is s0,n+…+sn,0. Now rn, or anything beyond rn, includes the first n rows. The positive terms sum to at least p-ε and the negative terms sum to at most q+ε. Together, rn is within ε of p+q. Therefore r approaches p+q.

If s is complex, it has a sum iff the real components have a sum and the imaginary components have a sum. There are positive and negative real components and positive and negative imaginary components. We can still add enough terms together, in our antidiagonal fashion, so that the positive reals add up to pre-ε, the negative reals add up to qre+ε, the positive imaginaries add up to pim-ε, and the negative imaginaries add up to qim+ε. Beyond n, the reals are within ε of their limit and the imaginaries are within ε of their limit. This generalizes to finite dimensional vectors.

If the terms of s are nonnegative, and the sum of r exists, add the elements of s in the antidiagonal order. Each partial sum is bounded by rn, which is below the limit of r, hence s is absolutely convergent, its sum is well defined, and by the above, its sum is given by the limit of r.

All this can be extended to 3 dimensions. Again, the sum exists only if s is an absolutely convergent series. Add terms in a diagonalized fashion, starting at the corner s0,0,0 and moving down and away from 0, or add up the terms in each 2 dimensional layer and then add these intermediate results. The two sums are equal, and the proof is essentially the same as that given above. Of course the layers can be added by rows and then columns, or columns and rows. We can add along the 3 dimensions in any of the 6 permutations; the result is the same.

Use induction to make the same statements about higher dimensional arrays. I'll leave the details to you.