Pathological Double Sums

We wish to know the total generated. Of course, we simply have to add the 12 numbers, butwe need to do thiswithout error. One way to check our answer is to add across all the rows and down all the columns. Now if we add the right-hand column and the bottom row of the extended array, we should obtain the same answer (it is 321). If we obtain the same answer we probably have the right answer; if we do not we have made an error. This seems obvious and trivial, but is it always true? Consider now a loan company with the capacity to offer an infinite number of loans over an infinite period. At the beginning of year 1 it loans, at no interest, a customer 1 unit of money. The customer repays half the loan (i.e. 2 unit) after 1 year, half the remainder after another year, and so on, repaying 1 2 unit, 1 4 unit, 1 8 unit etc. in successive years. Since 1 2 + 1 4 + 1 8 + · · · = 1, the loan will be recovered. However, at the end of the first year the company takes on another client on the same terms. Again the loan will be recovered. Every year the company takes on a new customer and evidently, in the limit, the company recovers all itsmoney and its deficit is 0. Now think of it another way. At the start the company is owed 1 unit. At the start of the second year it has loaned another unit and recovered 2 unit, so for that year its deficit is 1 + 2 units. Continuing in this way, its total deficit is 1 + 2 + 1 4 + 1 8 + · · · = 2, and not 0 as argued before. We set this out as in table 2, with negative signs denoting money lent. Adding all the row sums gives a total of 0 + 0+ 0+ · · · = 0, whereas adding the column sums gives a total of −1− 2 − 4 − 8 = −2. This is an example, inspired by an exercise on double integrals (ref. 1), where ∑∞ i=1 ∑∞ j=1 aij = ∑∞ j=1 ∑∞ i=1 aij , i.e. the order in which two infinite sums are carried out may not in general be interchanged.