# Summation machines and divergent series

December 7, 2016. We list a few desiderata for toting up an infinite list of numbers, and use these to get the “morally correct” answer for some famous divergent series. Extension material for real analysis.

### Introduction

Prerequisites: basic calculus and real analysis.

A few months ago, I watched some of Carl Bender’s excellent Perimeter lectures on mathematical physics. One of the pedagogical highlights was a simple method for generalising the textbook definition of convergence. Basically, start by writing down a few simple formal properties of infinite sums, and call anything that satisfies these properties a summation machine. This includes the usual $\epsilon$-$\delta$ notion of convergence of partial sums, but other things as well. In particular, you can use the assumed properties of a summation machine to deduce the formal value of certain famous divergent series (such as $1 - 1 + 1 - 1 \pm \cdots$ and $1 + 2 + 3 + \cdots$). This is a simple way to make the heuristic arguments of Euler and Ramanujan (and latter-day popularisers like Numberphile) somewhat more rigorous. We can summarise it easily:

A summation machine is a (partial) map from sequences to numbers which is additive and linear (pictured above). Of course, there are more rigorous ways to understand divergent series, but this is a nice way to formally capture some of the moves made in the non-rigorous derivations.

### What is a summation machine?

A summation machine is an operator $\mathcal{S}$ which takes a formal sequence of real numbers

$(a_n) := (a_1, a_2, a_3, a_4, \ldots), \quad a_n \in \mathbb{R},$

and either (a) spits out a finite real number, or (b) does not give an answer. (Technically, this makes it a partial function.) In the first case, we say that $(a_n)$ converges according to $\mathcal{S}$ or is $\mathcal{S}$-convergent. We call the number it spits out the sum according to $\mathcal{S}$ or $\mathcal{S}$-sum, and write

$\mathcal{S}(a_n) := \sum_{n\geq 1}^\mathcal{S} a_n.$

In case (b), we say the series is $\mathcal{S}$-divergent. We impose two further properties on $\mathcal{S}$, which generalise the behaviour of finite sums. Suppose $(a_n)$ and $(b_n)$ converge under $\mathcal{S}$. Then we must have

\begin{align*} \sum_{n\geq 1}^\mathcal{S} a_n & = a_1 + \sum_{n\geq 2}^\mathcal{S} a_n & \text{(additivity)}\\ \sum_{n\geq 1}^\mathcal{S} (\alpha a_n + \beta b_n) & = \alpha\sum_{n\geq 1}^\mathcal{S} a_n + \beta\sum_{n\geq 1}^\mathcal{S} b_n . & \text{(linearity)} \end{align*}

The first property forces $\mathcal{S}$ to behave like a sum. Thinking of a sequence to be summed as a stack, we can pop off the top item, process the tail of the stack, and add the top item to the result. This is not enough to nail down a notion of convergence, since we can choose to deal with the tail in different ways.

The second property forces $\mathcal{S}$ to obey some simple limit algebra. Another way to view the second property is the statement that summation machines $\mathcal{S}$ are linear transformations (in fact, linear functionals) on the space of real sequences, considered as vectors with an infinite number of components. (Requiring $\mathcal{S}$-convergence complicates this statement a little, but it is morally correct.)

Exercise. (a) Using induction, show that applying a summation machine to a finite sequence $(a_n)$ (that is, $a_n = 0$ for all $n$ bigger than some $k$) just gives the usual finite sum. (b) Show that additivity and linearity are equivalent to a single property:

$\sum_{n\geq 1}^\mathcal{S} (\alpha a_n + \beta b_n) = (\alpha a_1 + \beta b_2) + \alpha\sum_{n\geq 2}^\mathcal{S} a_n + \beta\sum_{n\geq 2}^\mathcal{S} b_n.$

(c) Give an example of a linear functional on the space of real sequences which is (i) linear but not additive; (ii) additive but not linear.

### Cesàro and Borel summation

Let’s give a couple of examples of summation machines. The first is just the “textbook” notion of the limit of partial sums:

$\mathcal{T}(a_n) := \lim_{N\to\infty}\sum_{n=1}^N a_n := \lim_{N\to\infty} A_N$

where we introduce the notation $A_N$ for the $N$-th partial sum. It’s not hard to see that this is additive and linear; we check the first property and leave the second as an (easy) exercise. From limit laws, we have

$\mathcal{T}(a_n) = \lim_{N\to\infty}\sum_{n=1}^N a_n = \lim_{N\to\infty}\left(a_1 + \sum_{n=2}^N a_n\right) = a_1 + \lim_{N\to\infty}\sum_{n=1}^{N-1} a_{n+1}.$

This is what was required. This notion of convergence is useful and historically important, but somewhat rigid.

A more flexible method is Cesàro summation $\mathcal{C}$. The Cesàro sum is just the average of the first $k$ partial sums as $k \to \infty$:

$\mathcal{C}( a_n) \equiv \lim_{k\to\infty} \frac{1}{k}\sum_{N=1}^k A_N,$

Clearly this is a map of the right sort. Linearity just follows from the linearity of limits and partial sums, so let’s check additivity:

\begin{align*} \mathcal{C}( a_n) & = \lim_{k\to\infty} \frac{1}{k}\sum_{N=1}^k A_N\\ & = \lim_{k\to\infty} \frac{1}{k}\sum_{N=1}^k \left(A_N - a_1 + a_1\right)\\ & = \lim_{k\to\infty} \frac{1}{k}\left(ka_1 + \sum_{N=1}^k \left(A_N - a_1\right)\right)\\ & = a_1 + \lim_{k\to\infty} \frac{1}{k}\sum_{N=1}^k A_N' = a_1 + \mathcal{C}(a_{n+1}), \end{align*}

where $A_N’$ denotes the $N$-th partial sum with respect to the modified sequence $a_2, a_3, \ldots$. Done!

Exercise. (a) Define Borel summation by

$\mathcal{B}(a_n) := \int_0^\infty \sum_{n=0}^\infty dt \,e^{-t} \frac{a_n t^n}{n!}.$

Show that Borel summation is linear and additive. (c) Is Grandi’s series (see below) $\mathcal{B}$-convergent?

### Some divergent series

At this point, we can apply these notions to some famous divergent series. First, Grandi’s series, where we must sum the sequence $a_n = (-1)^{n+1}$. It’s not hard to show that it is Cesàro-convergent.

Exercise.. Prove Grandi’s series has Cesàro of $1/2$.

Let us recapitulate the heuristic argument using summation machines. Suppose that $a_n$ is $\mathcal{S}$-convergent for a summation machine $\mathcal{S}$, with $s := \mathcal{S}(a_n)$. Then, from additivity and linearity,

$1 - s = 1 - \sum_{n\geq 1}^\mathcal{S} (-1)^{n+1} = 1 - 1 + \sum_{n\geq 2}^\mathcal{S} (-1)^{n} = \sum_{n\geq 1}^\mathcal{S} (-1)^{n+1} = s.$

Thus, $s = 1/2$ for any summation machine $\mathcal{S}$ for which $a_n$ is $\mathcal{S}$-convergent. This uses the same reasoning as the heuristic proof, but is a general result about notions of convergence! I think this is pretty neat. We let the reader prove a similar statement about the sum of natural numbers as an exercise.

Exercise (sum of natural numbers). (a) Consider the alternating sum of natural numbers,

$1 - 2 + 3 - 4 + \cdots +(-1)^{n+1}n + \cdots$

Assuming this sum and Grandi’s series are $\mathcal{S}$-convergent, show that

$\sum_{n\geq 1}^\mathcal{S} (-1)^{n+1} n = \frac{1}{4}.$

(b) Deduce that

$\sum_{n\geq 1}^ \mathcal{S} n = -\frac{1}{12}.$

Hint: Use part (a), linearity, and $L - 4L = -3L$, where $L$ is the limit.

### Conclusion

There are various questions we can ask about this basic setup. Can we always find a way to “tame the tail” of a given divergent series? We gave the example of Grandi’s series and Cesàro summation above, but for the sum of natural numbers, there is (as far as I know) no summation machine which gives the “correct” answer $-1/12$. If we relax our restrictions on summation methods, we can obtain this result using Euler-Maclaurin regularisation and related objects like the Riemann zeta function. Still, from a pedagogical standpoint, summation machines are a nice entrée to divergent series, encouraging us to think a little more deeply about the algebraic properties used in heuristic proofs.

Written on December 7, 2016