College Math Teaching

January 26, 2016

More Fun with Divergent Series: redefining series convergence (Cesàro, etc.)

Filed under: analysis, calculus, sequences, series — Tags: , , — collegemathteaching @ 10:21 pm

This post is more designed to entertain myself than anything else. This builds up a previous post which talks about deleting enough terms from a divergent series to make it a convergent one.

This post is inspired by Chapter 8 of Konrad Knopp’s classic Theory and Application of Infinite Series. The title of the chapter is Divergent Series.

Notation: when I talk about a series converging, I mean “converging” in the usual sense; e. g. if s_n = \sum_{k=0}^{k=n} a_k and lim_{n \rightarrow \infty}s_n = s then \sum_{k=0}^{\infty} a_k is said to be convergent with sum s .

All of this makes sense since things like limits are carefully defined. But as Knopp points out, in the “days of old”, mathematicians say these series as formal objects rather than the result of careful construction. So some of these mathematicians (like Euler) had no problem saying things like \sum^{\infty}_{k=0} (-1)^k = 1-1+1-1+1..... = \frac{1}{2} . Now this is complete nonsense by our usual modern definition. But we might note that \frac{1}{1-x} = \sum^{\infty}_{k=0} x^k for -1 < x < 1 and note that x = -1 IS in the domain of the left hand side.

So, is there a way of redefining the meaning of “infinite sum” that gives us this result, while not changing the value of convergent series (defined in the standard way)? As Knopp points out in his book, the answer is “yes” and he describes several definitions of summation that

1. Do not change the value of an infinite sum that converges in the traditional sense and
2. Allows for more series to coverge.

We’ll discuss one of these methods, commonly referred to as Cesàro summation. There are ways to generalize this.

How this came about

Consider the Euler example: 1 -1 + 1 -1 + 1 -1...... . Clearly, s_{2k} = 1, s_{2k+1} = 0 and so this geometric series diverges. But notice that the arithmetic average of the partial sums, computed as c_n = \frac{s_0 + s_1 +...+s_n}{n+1} does tend to \frac{1}{2} as n tends to infinity: c_{2n} = \frac{\frac{2n}{2}}{2n+1} = \frac{n}{2n+1} whereas c_{2n+1} = \frac{\frac{2n}{2}}{2n+2} =\frac{n}{2n+2} and both of these quantities tend to \frac{1}{2} as n tends to infinity.

So, we need to see that this method of summing is workable; that is, do infinite sums that converge in the previous sense still converge to the same number with this method?

The answer is, of course, yes. Here is how to see this: Let x_n be a sequence that converges to zero. Then for any \epsilon > 0 we can find M such that k > M implies that |x_k| < \epsilon . So for n > k we have \frac{x_1 + x_2 + ...+ x_{k-1} + x_k + ...+ x_n}{n} = \frac{x_1+ ...+x_{k-1}}{n} + \frac{x_k + x_{k+1} + ....x_n}{n} Because k is fixed, the first fraction tends to zero as n tends to infinity. The second fraction is smaller than \epsilon in absolute value. But \epsilon is arbitrary, hence this arithmetic average of this null sequence is itself a null sequence.

Now let x_n \rightarrow L and let c_n = \frac{x_1 + x_2 + ...+ x_{k-1} + x_k + ...+ x_n}{n} Now subtract note c_n-L = \frac{(x_1-L) + (x_2-L) + ...+ (x_{k-1}-L) +(x_k-L) + ...+ (x_n-L)}{n} and the x_n-L forms a null sequence. Then so do the c_n-L .

Now to be useful, we’d have to show that series that are summable in the Cesàro obey things like the multiplicative laws; they do but I am too lazy to show that. See the Knopp book.

I will mention a couple of interesting (to me) things though. Neither is really profound.

1. If a series diverges to infinity (that is, if for any positive M there exists n such that for all k \geq n, s_k > M , then this series is NOT Cesàro summable. It is relatively easy to see why: given such an M, k then consider \frac{s_1 + s_2 + s_3 + ...+s_{k-1} + s_k + s_{k+1} + ...s_n}{n} = \frac{s_1+ s_2 + ...+s_{k-1}}{n} + \frac{s_k + s_{k+1} .....+s_{n}}{n} which is greater than \frac{n-k}{n} M for large n . Hence the Cesàro partial sum becomes unbounded.

Upshot: there is no hope in making something like \sum^{\infty}_{n=1} \frac{1}{n} into a convergent series by this method. Now there is a way of making an alternating, divergent series into a convergent one via doing something like a “double Cesàro sum” (take arithmetic averages of the arithmetic averages) but that is a topic for another post.

2. Cesàro summation may speed up convergent of an alternating series which passes the alternating series test, OR it might slow it down. I’ll have to develop this idea more fully. But I invite the reader to try Cesàro summation for \sum^{\infty}_{k=1} (-1)^{k+1} \frac{1}{k} and on \sum^{\infty}_{k=1} (-1)^{k+1} \frac{1}{k^2} and on \sum^{\infty}_{k=0} (-1)^k \frac{1}{2^k} . In the first two cases, the series converges slowly enough so that Cesàro summation speeds up convergence. Cesàro slows down the convergence in the geometric series though. It is interesting to ponder why.

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