As you recall, each series implies a sequence of partial sums. Let u hold the partial sums of s and let v hold the partial sums of t. Thus un is the sum of the first n terms of s, and vn is the sum of the first n terms of t.

Let w be the term by term product of u and v. Thus wn is the first n terms of s times the first n terms of t. The limit of w is the limit of u times the limit of v, which is the sum of s times the sum of t. Let's evaluate the limit of w.

The nth term of w is the sum of sitj as i and j run from 1 to n. Let M be a 2 dimensional matrix with Mi,j = sitj. Now wn is the sum of the upper left entries of M, n rows and n columns. As n increases we cover the entire array. The limit of w is the limit of M, as a 2 dimensional series. As long as M converges, the limit of w is the limit of s times the limit of t, which is what we want to prove.

The terms of M are nonnegative, so we can add across rows, and then add the rows together. If this nested sum converges then M converges.

Let q be the sum over t. Thus the kth row of M sums to sk×q. Add up all the rows, and we are adding the series s, scaled by q. This is convergent, and its limit is the sum of s multiplied by q, or the sum of s times the sum of t.

either s or t could be a nonpositive series, instead of a nonnegative series; the theorem still holds. As long as the entries of M have the same sign, positive or negative, we have absolute convergence, and the sum of M is the limit of w, is the sum of s*t.

Next let s and t be absolutely convergent series. Each series could contain both positive and negative terms. Our proof runs into a snag when we evaluate the sum of M. If has positive and negative terms; does it converge? Write M as the sum of two matrices, one with the positive terms and one with the negative terms. Each is dominated by a matrix where all the terms of s and t are made positive, and that matrix converges. Thus the positive and negative "halves" of M are absolutely convergent, and M is absolutely convergent. Once again the product s*t converges to the sum of s times the sum of t.

Use induction to generalize this to a finite product of absolutely convergent series. For instance, the product of the three series s t and u is the sum of sitjuk for all i j and k > 0, and the limit of this 3 dimensional series is equal to the sum of s times the sum of t times the sum of u.

Show that series multiplication is commutative and associative. Furthermore, multiplication distributes over addition. In other words, (s+t)*u = s*u+t*u.