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.

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.

January 18, 2014

Fun with divergent series (and uses: e. g. string theory)

One “fun” math book is Knopp’s book Theory and Application of Infinite Series. I highly recommend it to anyone who frequently teaches calculus, or to talented, motivated calculus students.

One of the more interesting chapters in the book is on “divergent series”. If that sounds boring consider the following:

we all know that $\sum^{\infty}_{n=0} x^n = \frac{1}{1-x}$ when $|x| < 1$ and diverges elsewhere, PROVIDED one uses the “sequence of partial sums” definition of covergence of sums. But, as Knopp points out, there are other definitions of convergence which leaves all the convergent (by the usual definition) series convergent (to the same value) but also allows one to declare a larger set of series to be convergent.

Consider $1 - 1 + 1 -1 + 1.......$

of course this is a divergent geometric series by the usual definition. But note that if one uses the geometric series formula:

$\sum^{\infty}_{n=0} x^n = \frac{1}{1-x}$ and substitutes $x = -1$ which IS in the domain of the right hand side (but NOT in the interval of convergence in the left hand side) one obtains $1 -1 +1 -1 + 1.... = \frac{1}{2}$.

Now this is nonsense unless we use a different definition of sum convergence, such as the Cesaro summation: if $s_k$ is the usual “partial sum of the first $k$ terms: $s_k = \sum^{n=k}_{n =0}a_n$ then one declares the Cesaro sum of the series to be $lim_{m \rightarrow \infty} \frac{1}{m}\sum^{m}_{k=1}s_k$ provided this limit exists (this is the arithmetic average of the partial sums).

(see here)

So for our $1 -1 + 1 -1 ....$ we easily see that $s_{2k+1} = 0, s_{2k} = 1$ so for $m$ even we see $\frac{1}{m}\sum^{m}_{k=1}s_k = \frac{\frac{m}{2}}{m} = \frac{1}{2}$ and for $m$ odd we get $\frac{\frac{m-1}{2}}{m}$ which tends to $\frac{1}{2}$ as $m$ tends to infinity.

Now, we have this weird type of assignment.

But that won’t help with $\sum^{\infty}_{k = 1} k = 1 + 2 + 3 + 4 + 5.....$. But weirdly enough, string theorists find a way to assign this particular series a number! In fact, the number that they assign to this makes no sense at all: $-\frac{1}{12}$.

What the heck? Well, one way this is done is explained here:

Consider $\sum^{\infty}_{k=0}x^k = \frac{1}{1-x}$ Now differentiate term by term to get $1 +2x + 3x^2+4x^3 .... = \frac{1}{(1-x)^2}$ and now multiply both sides by $x$ to obtain $x + 2x^2 + 3x^3 + .... = \frac{x}{(1-x)^2}$ This has a pole of order 2 at $x = 1$. But now substitute $x = e^h$ and calculate the Laurent series about $h = 0$; the 0 order term turns out to be $\frac{1}{12}$. Yes, this has applications in string theory!

Now of course, if one uses the usual definitions of convergence, I played fast and loose with the usual intervals of convergence and when I could differentiate term by term. This theory is NOT the usual calculus theory.

Now if you want to see some “fun nonsense” applied to this (spot how many “errors” are made….it is a nice exercise):

What is going on: when one sums a series, one is really “assigning a value” to an object; think of this as a type of morphism of the set of series to the set of numbers. The usual definition of “sum of a series” is an especially nice morphism as it allows, WITH PRECAUTIONS, some nice algebraic operations in the domain (the set of series) to be carried over into the range. I say “with precautions” because of things like the following:

1. If one is talking about series of numbers, then one must have an absolutely convergent series for derangements of a given series to be assigned the same number. Example: it is well known that a conditionally convergent alternating series can be arranged to converge to any value of choice.

2. If one is talking about a series of functions (say, power series where one sums things like $x^n$) one has to be in OPEN interval of absolute convergence to justify term by term differentiation and integration; then of course a series is assigned a function rather than a number.

So when one tries to go with a different notion of convergence, one must be extra cautious as to which operations in the domain space carry through under the “assignment morphism” and what the “equivalence classes” of a given series are (e. g. can a series be deranged and keep the same sum?)