Approximaths

Borel summation

Written by

approximaths

in

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.

Comments

Leave a comment

More posts