Let s be a series of complex numbers, its partial sums bounded by the radius r. Let t be a sequence of monotonically decreasing real numbers, approaching 0. This does not mean t, as a series, is convergent, and s certainly need not be convergent. We want to know if the series sjtj converges.

Let w be the sequence of partial sums derived from s. In other words, wn = s1 + s2 + s3 + … + sn, and wn is bounded by r.

Start with wn×tn+1. Then add wn×(tn-tn+1). This leaves wntn. Then add wn-1×(tn-1-tn). This produces sntn + wn-1tn-1. Then add wn-2×(tn-2-tn-1). This produces sntn + sn-1tn-1 + wn-2tn-2. Well you can probably see the pattern. Continue adding terms down to w1×(t1-t2) and get the following equation for the partial sum of sjtj from 1 to n.

s1t1 + s2t2 + s3t3 + … + sntn =

wntn+1 +

w1(t1-t2) + w2(t2-t3) + w3(t3-t4) + … + wn(tn-tn+1)

We can now test for convergence by examining the right hand side. The term wntn+1 has a norm that is bounded by rtn+1, which approaches 0. Thus this sequence approaches 0, though it may do so in a rather chaotic manner.

We only need consider the last series. Replace wi with r, and the new series r×(tn-tn+1) is convergent, because it telescopes. The terms of this series are all positive, hence it converges absolutely. And it dominates the norms of the series wn×(tn-tn+1), so that series also converges absolutely. This means the series sjtj is convergent.

In summary, the series sjtj converges if the (complex) partial sums of s are bounded, and the real sequence t decreases to 0. Note that s could be any finite dimensional vector sequence, but it usually contains complex numbers, the two dimensional case.

If z is a point on or inside the unit circle, other than 1, and t is a sequence of real numbers that monotonically decreases to 0, the series zjtj converges. Consider the partial sum 1+z+z2+z3+…+zn. This is zn+1-1 over z-1, which has a radius no larger than 2/|(z-1)|. The conditions of the theorem have been met. For a given z, we may not know what the sum of zn/n is, but it does converge.

Note that the alternating series is a special case of the above, with z set to -1.

As you recall, the term wntn+1 had a norm bounded by rtn+1. Given ε, find k such that tk is less than ½ε/r. Thus wntn+1 is bounded by half of ε.

Next, turn to the telescoping series. Its error term is bounded by rtn. Beyond k, this too is less than half of ε. Put this together, and all the partial sums of sjtj, beyond k, are within ε of the sum.

Given the sequence t, and the radius r, the speed of convergence can be described, in a manner that does not depend on s.

If t is itself a convergent series, such as a geometric series, it is absolutely convergent, its terms being positive. The partial sums of s are bounded by r, hence the terms of s are bounded by 2r. The terms of sjtj are bounded below an absolutely convergent series, and abel summation implies absolute convergence.

bound each wn with r, as we did before. Of course wntn+1 still approaches 0.

With each wi replaced with r, the telescoping series converges to rt1. Once we have skipped past a finite number of terms, all the terms lie in the first quadrant. The series is absolutely convergent in its real and imaginary components, hence it is absolutely convergent. This means the original, telescoping series was absolutely convergent, and the entire series converges.

Once again the speed of convergence is set by r and t, using the proof outlined above.