Category: Summation techniques

Borel summation

Let introduce the Borel summation by first recalling the following property:

$\displaystyle n!= \int_{0}^{\infty} e^{-t}t^{n}dt$

$\displaystyle 1 = \frac{\int_{0}^{\infty} e^{-t}t^{n}dt}{n!}$

We would like to sum the following series:

$\displaystyle \sum_{n=0}^{\infty}a_n = a_0 + a_1 + a_2 + a_3 + a_4 +...$
$\displaystyle \phantom{\sum_{n=0}^{\infty}a_n} = 1a_0 + 1a_1 + 1a_2 + 1a_3 + 1a_4 +...$
$\displaystyle \phantom{\sum_{n=0}^{\infty}a_n} = \sum_{n=0}^{\infty} \frac{\int_{0}^{\infty} e^{-t}t^{n}dt}{n!} a_n$

The Borel sum $B$ is defined as follow:

$\displaystyle B := \sum_{n=0}^{\infty} \frac{\int_{0}^{\infty} e^{-t}t^{n}dt}{n!} a_n$
$\displaystyle \phantom{B :=} = \int_{0}^{\infty}e^{-t} \sum_{n=0}^{\infty} \frac{t^n}{n!}a_n\,dt$

The factorial term makes this sum with better chance to converge. In the section concerning the Euler summation we have seen that:

$\displaystyle E(1 - 1 + 1 - 1 + 1 - 1 + ...) = \frac{1}{2}$

Let calculate the corresponding Borel sum:

$\displaystyle B(1 - 1 + 1 - 1 + 1 - 1 + ...) = \int_{0}^{\infty}e^{-t} \sum_{n=0}^{\infty} \frac{t^n}{n!}(-1)^n \,dt$
$\displaystyle \phantom{B(1 - 1 + 1 - 1 + 1 - 1 + ...)} = \int_{0}^{\infty}e^{-t} \sum_{n=0}^{\infty} \frac{(-t)^n}{n!} \,dt$
$\displaystyle \phantom{B(1 - 1 + 1 - 1 + 1 - 1 + ...)} = \int_{0}^{\infty} e^{-t}e^{-t}\,dt$
$\displaystyle \phantom{B(1 - 1 + 1 - 1 + 1 - 1 + ...)} = \int_{0}^{\infty} e^{-2t}\,dt$
$\displaystyle \phantom{B(1 - 1 + 1 - 1 + 1 - 1 + ...)} = -\frac{1}{2} e^{-2t}\Big|_{0}^{\infty}$
$\displaystyle \phantom{B(1 - 1 + 1 - 1 + 1 - 1 + ...)} = 0 - (-\frac{1}{2})$
$\displaystyle \phantom{B(1 - 1 + 1 - 1 + 1 - 1 + ...)} = \frac{1}{2}$

and therefore:

$\displaystyle B(1-1+1-1+1-1+...) = E(1-1+1-1+1-1+...)$

Consider the series:

$\displaystyle -1+2!-3!+4!-5!+6!-...$

$\displaystyle B(-1+2!-3!+4!-...) = \int_{0}^{\infty}e^{-t} \sum_{n=1}^{\infty} \frac{t^n}{n!}(-1)^n n! \,dt$
$\displaystyle \phantom{B(-1+2!-3!+4!-...)} = \int_{0}^{\infty}e^{-t} \sum_{n=1}^{\infty} (-t)^n \,dt$

$\sum_{n=0}^{\infty} (-t)^n$ converges for $t < 1$ to $-\frac{1}{1+t}$ therefore:

$\displaystyle B = -\int_{0}^{\infty} \frac{te^{-t}}{(1+t)} \,dt \approx -0.40365$

If a series is summable in the sense of Euler, then it is also summable in the sense of Borel, and both summation methods yield the same value. The converse is false: there exist series that are summable in the sense of Borel but not in the sense of Euler. In other words, Borel summation is more powerful, as it applies to more strongly divergent series.

November 13, 2025
Euler summation II

In the previous post, we introduced Euler summation. The following are two examples where it fails to produce a finite result.

Consider the divergent series:

$\displaystyle 0 + 1 + 2 + 3 + 4 + \dots$

Define:

$\displaystyle \begin{array}{rcl} f(x) &=& 0x^{0} + 1x^{1} + 2x^{2} + 3x^{3} + 4x^{4} + \dots \\ &=& \sum_{n=0}^{\infty} nx^{n} = \frac{x}{(1-x)^{2}} \\ \end{array}$

The Euler sum is:

$\displaystyle E = \lim_{x \to 1_{-}} \frac{x}{(1-x)^{2}} = \infty$

$\displaystyle E(0+ 1 + 2 + 3 + 4 + \dots) = \infty$

Now consider the divergent series:

$\displaystyle 1 + 4 + 9 + 16 + 25 + 36 + \dots$

Define:

$\displaystyle \begin{array}{rcl} f(x) &=& 1^{2}x^{1}+2^{2}x^{2}+3^{2}x^{3}+4^{2}x^{4}+5^{2}x^{5}+6^{2}x^{6}+\dots \\ &=& \sum_{n=1}^{\infty} n^{2} x^{n} = \frac{x(1+x)}{(1-x)^{3}} \\ \end{array}$

The Euler sum is:

$\displaystyle E = \lim_{x \to 1_{-}}\frac{x(1+x)}{(1-x)^{3}} = \infty$

$\displaystyle E(1^{2} + 2^{2} + 3^{2} + 4^{2} + \dots) = \infty$

Advantages

Regularization of slowly divergent series:
Euler summation can assign a finite value to some divergent series that oscillate or diverge slowly, such as

$\displaystyle 1 - 1 + 1 - 1 + \dots$

where $E$ (series) = $\tfrac{1}{2}$ .

Improved convergence:
For many convergent series, Euler transformation accelerates convergence, making it useful for numerical computations.

Analytic continuation link:
It provides a bridge between ordinary summation and more advanced summation methods (e.g. Borel or zeta regularization).

Disadvantages

Limited domain of applicability:
Euler summation fails for series that diverge too rapidly, such as

$\displaystyle 1 + 2 + 3 + 4 + \dots$

where $E$ (series) = $\infty$ .

Not uniquely defined for all divergent series:
Some series cannot be assigned a finite Euler sum, or the method may yield inconsistent results depending on the transformation order.

Weaker than analytic regularization:
Compared to zeta or Borel summation, Euler’s method handles fewer classes of divergent series and lacks a rigorous analytic continuation framework.

November 6, 2025
Euler summation I

So far we’ve seen Padé’s approximants. These have enabled us to approximate a function from its corresponding Taylor series and then transform this (potentially non-convergent) series into a convergent rational fraction.

We’d like to introduce other techniques that could be used to sum non-convergent series. First, we’ll take a look at Euler summation.

If the series $\sum_{n=0}^{\infty} a_n$ is algebraically divergent (the terms blow up like some power of n), then the series:

$\displaystyle f(x) = \sum_{n=0}^{\infty}a_n x^n$

converges for $x \in (-1,1)$ . If the limit

$\displaystyle E := \lim_{x \to 1_{-}}f(x)$

exists and is finite then it is called the Euler sum $E$ of the original series.

For example, consider the divergent series:

$\displaystyle 1 - 1 + 1 - 1 + 1 - 1 + ...$

and multiply each n-term with $x^n$ (starting at n = 0):

$\displaystyle f(x) = x^0 - x^1 + x^2 - x^3 + x^4 - x^5 + ...$

$\displaystyle = \frac{1}{1-(-x)} = \frac{1}{1+x}$

$\displaystyle E = \lim_{x \to 1_{-}}{\frac{1}{1+x}} = \frac{1}{2}$

Therefore (according to Euler summation):

$\displaystyle 1 - 1 + 1 - 1 + 1 - 1 + ... = \frac{1}{2}$

October 30, 2025