# Borel summation

Borel, then an unknown young man, discovered that his summation method gave the 'right' answer for many classical divergent series. He decided to make a pilgrimage to Stockholm to see Mittag-Leffler, who was the recognized lord of complex analysis. Mittag-Leffler listened politely to what Borel had to say and then, placing his hand upon the complete works by Weierstrass, his teacher, he said in Latin, 'The Master forbids it'.

Mark Kac, quoted by Reed & Simon (1978, p. 38)

In mathematics, Borel summation is a summation method for divergent series, introduced by Émile Borel (1899). There are several variations of this method that are also called Borel summation, and a generalization of it called Mittag-Leffler summation.

## Definition

Throughout let A(z) denote a formal power series

$A(z) = \sum_{k = 0}^\infty a_kz^k$,

and define the Borel transform of A to be its equivalent exponential series

$\mathcal{B}A(t) \equiv \sum_{k=0}^\infty \frac{a_k}{k!}t^k.$

### Borel summation

Suppose that the Borel transform converges to an analytic function of t near 0 that can be analytically continued along the positive real axis to a function growing sufficiently slowly that the following integral is well defined (as an improper integral), the Borel sum of A is given by

$\int_0^\infty e^{-t} \mathcal{B}A(tz) \, dt.$

If the integral converges at z ∈ C to some a(z), we say that the Borel sum of A converges at z, and write ${\textstyle \sum} a_kz^k = a(z) \,(\boldsymbol B)$.

### Weak Borel summation

Let An(z) denote the partial sum

$A_n(z) = \sum_{k=0}^n a_k z^k.$

A weaker form of Borel's summation method defines the Borel sum of A to be

$\lim_{t\rightarrow\infty} e^{-t}\sum_{n=0}^\infty \frac{t^n}{n!}A_n(z).$

If this converges at z ∈ C to some a(z), we say that the weak Borel sum of A converges at z, and write ${\textstyle \sum} a_kz^k = a(z) \, (\boldsymbol{wB})$.

## Basic properties

### Regularity

The methods (B) and (wB) are both regular summation methods, meaning that whenever A(z) converges (in the standard sense), then the Borel sum and weak Borel sum also converge, and do so to the same value. i.e.

$\sum_{k=0}^\infty a_k z^k = A(z) < \infty \quad \Rightarrow \quad {\textstyle \sum} a_kz^k = A(z) \,\, (\boldsymbol{B},\,\boldsymbol{wB}).$

Regularity of (B) is easily seen by a change in order of integration: if A(z) is convergent at z,then

$A(z) = \sum_{k=0}^\infty a_k z^k = \sum_{k=0}^\infty a^k \left( \int_{0}^\infty e^{-t}t^k dt \right) \frac{z^k}{k!} = \int_{0}^\infty e^{-t} \sum_{k=0}^\infty a_k \frac{(tz)^k}{k!}dt,$

where the rightmost expression is exactly the Borel sum at z.

Regularity of (B) and (wB) imply that these methods provide analytic extensions to A(z).

### Equivalence of Borel and weak Borel summation

Any series A(z) that is weak Borel summable at z ∈ C is also Borel summable at z. However, one can construct examples of series which are divergent under weak Borel summation, but which are Borel summable. The following theorem characterises the equivalence of the two methods.

Theorem ((Hardy 1992, 8.5)).
Let A(z) be a formal power series, and fix z ∈ C, then:
1. If ${\textstyle \sum} a_kz^k = a(z) \, (\boldsymbol{wB})$, then ${\textstyle \sum}a_kz^k = a(z) \, (\boldsymbol{B})$.
2. If ${\textstyle \sum} a_kz^k = a(z) \, (\boldsymbol{B})$, and $\lim_{t \rightarrow \infty} e^{-t}\mathcal B A(zt) = 0,$ then ${\textstyle \sum} a_kz^k = a(z) \, (\boldsymbol{wB})$.

### Relationship to other summation methods

• (B) is the special case of Mittag-Leffler summation with α = 1.
• (wB) can be seen as the limiting case of generalized Euler summation method (E,q) in the sense that as q → ∞ the domain of convergence of the (E,q) method converges up to the domain of convergence for (B).[1]

## Examples

### The geometric series

Consider the geometric series

$A(z) = \sum_{k = 0}^\infty z^k,$

which converges (in the standard sense) to 1/(1 − z) for |z| < 1. The Borel transform is

$\mathcal{B}A(t) \equiv \sum_{k=0}^\infty \frac{1}{k!}t^k = e^t,$

from which we obtain the Borel sum

$\int_0^\infty e^{-t}\mathcal{B}A(tz) \, dt = \int_0^\infty e^{-t} e^{tz} \, dt =\frac{1}{1-z}$

which converges in the larger region Re(z) < 1, giving an analytic continuation of the original series.

Considering instead the weak Borel transform, the partial sums are given by AN(z) = (1-zN+1)/(1-z), and so the weak Borel sum is

$\lim_{t \rightarrow \infty}e^{-t} \sum_{n=0}^\infty \frac{1 -z^{n+1}}{1-z} \frac{t^n}{n!} = \lim_{t \rightarrow \infty} \frac{e^{-t}}{1-z} \big( e^t - z e^{tz} \big) = \frac{1}{1-z},$

where, again, convergence is on Re(z) < 1. Alternatively this can be seen by appealing to part 2 of the equivalence theorem, since for Re(z) < 1

$\lim_{t \rightarrow \infty} e^{-t} (\mathcal{B} A)(zt) = e^{t(z-1)} = 0.$

### An alternating factorial series

Consider the series

$A(z) = \sum_{k = 0}^\infty k!\left(-1 \cdot z\right)^k,$

then A(z) does not converge for any nonzero z ∈  C. The Borel transform is

$\mathcal{B}A(t) \equiv \sum_{k=0}^\infty \left(-1 \cdot t\right)^k = \frac{1}{1+t}$

for |t| < 1, which can be analytically continued to all t ≥ 0. So the Borel sum is

$\int_0^\infty e^{-t}\mathcal{B}A(tz) \, dt = \int_0^\infty \frac{e^{-t}} {1+tz} \, dt = \frac 1 z \cdot e^\frac 1 z \cdot \Gamma\left(0,\frac 1 z \right)$

(where Γ is the incomplete Gamma function).

This integral converges for all z ≥ 0, so the original divergent series is Borel summable for all such z. This function has an asymptotic expansion as z tends to 0 that is given by the original divergent series. This is a typical example of the fact that Borel summation will sometimes "correctly" sum divergent asymptotic expansions.

Again, since

$\lim_{t \rightarrow \infty}e^{-t} (\mathcal B A)(zt) = \lim_{t \rightarrow \infty} \frac{e^{-t}}{1 + zt} = 0,$

for all z, the equivalence theorem ensures that weak Borel summation has the same domain of convergence, z ≥ 0.

### An example in which equivalence fails

The following example extends on that given in (Hardy 1992, 8.5). Consider

$A(z) = \sum_{k = 0}^\infty \left( \sum_{l=0}^\infty \frac{(-1)^l(2l + 2)^k}{(2l+1)!} \right) z^k.$

After changing the order of summation, the Borel transform is given by

\begin{align} \mathcal B A(t)&= \sum_{l = 0}^\infty \left( \sum_{k=0}^\infty \frac{\big((2l+2) t\big)^k}{k!} \right) \frac{(-1)^l}{(2l+1)!} \\ &= \sum_{l=0}^\infty e^{(2l+2)t}\frac{(-1)^l}{(2l+1)!} \\ &= e^t \sum_{l=0}^\infty \big(e^t\big)^{2l+1} \frac{(-1)^l}{(2l+1)!} \\ & = e^t \sin\left( e^t \right). \end{align}

At z = 2 the Borel sum is given by

$\int_{0}^\infty e^t \sin(e^{2t})dt = \int_{1}^\infty \sin(u^2)du = \frac{\sqrt{\pi}}{8} - S(1) < \infty,$

where S(x) is the Fresnel integral. Via the convergence theorem along chords, the Borel integral converges for all z ≤ 2 (clearly the integral diverges for z > 2).

For the weak Borel sum we note that

$\lim_{t \rightarrow \infty} e^{(z-1)t}\sin \left( e^{zt} \right) = 0$

holds only for z < 1, and so the weak Borel sum converges on this smaller domain.

## Existence results and the domain of convergence

### Summability on chords

If a formal series A(z) is Borel summable at z0 ∈ C, then it is also Borel summable at all points on the chord Oz0 connecting z0 to the origin. Moreover, there exists a function a(z) analytic throughout the disk with radius Oz0 such that

${\textstyle \sum} a_kz^k = a(z) \, (\boldsymbol B),$

for all z = θz0, θ ∈ [0,1].

An immediate consequence is that the domain of convergence of the Borel sum is a star domain in C. More can be said about the domain of convergence of the Borel sum, than that it is a star domain, which is referred to as the Borel polygon, and is determined by the singularities of the series A(z).

### The Borel polygon

Suppose that A(z) has strictly positive radius of convergence, so that it is analytic in a non-trivial region containing the origin, and let SA denote the set of singularities of A. This means that P ∈ SA if and only if A can be continued analytically along the open chord from 0 to P, but not to P itself. For P ∈ SA, let LP denote the line passing through P which is perpendicular to the chord OP. Define the sets

$\Pi_P = \{z \in \mathbb{C} \, \colon \, Oz \cap L_P = \varnothing \},$

the set of points which lie on the same side of LP as the origin. The Borel polygon of A is the set

$\Pi_A = \text{cl}\Big( \bigcap_{P \in S_A} \Pi_P \Big).$

An alternative definition was used by Borel and Phragmén (Sansone & Gerretsen 1960, 8.3). Let $S \subset \mathbb{C}$ denote the largest star domain on which there is an analytic extension of A, then $\Pi_A$ is the largest subset of $S$ such that for all $P \in \Pi_A$ the interior of the circle with diameter OP is contained in $S$. Referring to the set $\Pi_A$ as a polygon is somewhat of a misnomer, since the set need not be polygonal at all; if, however, A(z) has only finitely many singularities then $\Pi_A$ will in fact be a polygon.

The following theorem, due to Borel and Phragmén provides convergence criteria for Borel summation.

Theorem (Hardy 1992, 8.8).
The series A(z) is (B) summable at all $z \in \text{int}(\Pi_A)$, and is (B) divergent at all $z \in \mathbb{C}\backslash \Pi_A$.

Note that (B) summability for $z \in \partial \Pi_A$ depends on the nature of the point.

#### Example 1

Let ωi ∈ C denote the m-th roots of unity, i =1,...m, and consider

\begin{align} A(z) & = \sum_{k=0}^\infty (\omega_1^k + \ldots + \omega_m^k)z^k \\ & = \sum_{i=1}^m \frac{1}{1-\omega_iz}, \end{align}

which converges on B(0,1) ⊂ C. Seen as a function on C, A(z) has singularities at SA = {ωi : i = 1,...m}, and consequently the Borel polygon $\Pi_A$ is given by the regular m-gon centred at the origin, and such that 1 ∈ C is a midpoint of an edge.

#### Example 2

The formal series

$A(z) = \sum_{k=0}^\infty z^{2^k},$

converges for all $|z| < 1$ (for instance, by the comparison test with the geometric series). It can however be shown[2] that A does not converge for all points z ∈ C with z2n = 1. Since the set of such z is dense in the unit circle, there can be no analytic extension of A outside of B(0,1). Subsequently the largest star domain to which A can be analytically extended is S = B(0,1) from which (via the second definition) one obtains $\Pi_A = B(0,1)$. In particular we see that the Borel polygon is in fact not polygonal.

### A Tauberian Theorem

A Tauberian theorem provides conditions under which convergence of one summation method, implies convergence under another method. The principal Tauberian theorem[1] for Borel summation provides conditions under which the weak Borel method implies convergence of the series.

Theorem (Hardy 1992, 9.13). :If A is (wB) summable at z0 ∈ C, ${\textstyle \sum}a_kz_0^k = a(z_0) \, (\boldsymbol{wB})$, and
$a_kz_0^k = O \left( k^{-{\textstyle \frac{1}{2}}} \right), \qquad \forall k \geq 0,$
then $\sum_{k=0}^\infty a_kz_0^k = a(z_0)$, and the series converges for all |z|<|z0|.

## Applications

Borel summation finds application in perturbation expansions in quantum field theory. In particular in 2-dimensional Euclidean field theory the Schwinger functions can often be recovered from their perturbation series using Borel summation (Glimm & Jaffe 1987, p. 461). Some of the singularities of the Borel transform are related to instantons and renormalons in quantum field theory (Weinberg 2005, 20.7).