# Summing Divergent Series and Borel Summation

• I
Mentor
I have recently been investigating summing divergent series and zeta function regularization's relation to dimensional re-normalization. Making some progress, but it is a bit slow despite literature being available:
https://www.imperial.ac.uk/media/im.../dissertations/2009/Nicolas-Robles-Thesis.pdf

Anyway part of this has been sorting out exactly what is going on with summing divergent series? I was particularly struck by a technique often used in normal QM - Borel Summation. As background see some stuff by Carl Bender:
https://arxiv.org/abs/1703.05164

So lets start at the beginning. Consider the sinc function sine(x)/x. Most would say it's continuous - but is it really? Well at 0 its 0/0 which is undefined - some books call it a singularity - but that's not the terminology I would use - simply undefined - but I will call it a singularity to conform with other sources. Using l'hopital its removable and its value at 0 is defined as 1 so its continuous. Most wouldn't even think about it - but strictly speaking its a removable singularity.

Here is an interesting question - if you can remove singularities can you remove divergences? The surprising answer is yes. In general here is how its done. Suppose ∑a(n)*x^n has some radius of convergence. Above that radius its divergent. But at least in some cases can those divergences be removed? Yes it can, by using what's called analytic continuation. If you don't know about this bit of magic here is a primer (including some information on Borel summation - but not at the level I would like):

I couldn't find any sources that satisfied me, so needed to figure it out myself.

Here is what's I nutted out - which could be wrong. Suppose you have ∑a(n) then Σa(n)*n!/n! = Σa(n)*(∫t^n*e^-t)/n!, using Γ(n+1) = n!. Lets formally interchange the integral and sum and you get ∑a(n) = ∫∑(a(n)/n!)*(t^n*e^-t). This is called the Borel sum. If limit n → ∞ |a(n+1)/a(n)| is < 1 ie ∑a(n) is absolutely convergent, then the power series ∑(a(n)/n!)*t^n has an infinite radius of convergence because of that n! in the denominator. This means ∑|a(n)|/n!)*(t^n*e^-t) < ∞ for all t, so is encouraging. The issue is reversing the order of sum and integral. You can consider the sum as an integral of a function where the value at n is f(n), where f(n) is a series. Then you can apply Fubini's Theorem to Σ∫|a(n)|*(t^n*e^-t)/n! < ∞. Its < ∞ because a(n) is absolutely convergent so the interchange is justified.

All right so what do we have. Consider the power series ∑a(n)*x^n with a radius of convergence <=1 which means it may be divergent at x = 1. But doing Borel summation we have ∫∑(a(n)/n!)*(xt)^n*e^-t) and equals ∑a(n)*x^n in its radius of convergence. This means you should be able to analytically continue it to x =1, and since analytic continuations are basically unique the Borel summation is the analytic continuation of ∑a(n)*x^n. So on setting x =1 we have ∑a(n) = ∫Σa(n)*(t^n)*e^-t)/n!.

So that's it - basically Borel Summation, and divergent summation in general, is just analytic continuation.

Now have to get back to sorting out dimensional re-normalization and zeta function regularization, but now I at least understand the general principle.

Thanks
Bill

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A. Neumaier
Then you can apply Fubini's Theorem
When the original series is divergent you cannot apply Fubini's theorem, since it asserts that both sides are meaningful beforehand.

The point of any summation method with some mathematical theory behind is that you assume some additional global mathematical property about the given function, e.g., an integral representation such as in the Borel summation formula. One first shows that this property implies that the function is infinitely often differentiable and hence has a Taylor series. Then one shows that the coefficients of the Taylor series determine the function completely. Thus having the coefficients one has the function, independent of any convergence statement for the Taylor series.

Of course any summation method is only as good for a particular application as the assumptions that go into it fit this application. This has to be so since the same formal power series is asymptotic to uncountably many different functions. For example, the formal power series with all coefficients zero is summed by any reasonable summation method (including Borel summation and zeta function regularization) to the zero function, although it also arises by Taylor expansion of the ##C^\infty## function defined for any ##a>0## by ##f(0)=0## and ##f(x)=e^{-a/x^2}## for ##x\ne0##.

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• bhobba
Mentor
When the original series is divergent you cannot apply Fubini's theorem, since it asserts that both sides are meaningful beforehand.

Yes but I assumed ∑a(n) was absolutely convergent so ∑|a(n)| = Σ∫|a(n)|*(t^n*e^-t)/n! < ∞ and Fubini is applicable. The rest of your post is course all true.

The deep answer of course is Watsons Lemma which I didn't want to get into due to the I nature of the thread.

BTW thanks for commenting - I am on my own here - the literature I have read didn't satisfy me and a mathematician looking at it is is very helpful.

Thanks
Bill

A. Neumaier
but I assumed ∑a(n) was absolutely convergent
The divergent series in quantum physics usually don't have this property; the convergence radius is usually zero.
basically Borel Summation, and divergent summation in general, is just analytic continuation.
If the convergence radius is zero your argument breaks down completely and you cannot analytically continue.

• bhobba
Mentor
The divergent series in quantum physics usually don't have this property; the convergence radius is usually zero. If the convergence radius is zero your argument breaks down completely and you cannot analytically continue.

Drat's - but you saved me a lot of work.

Thanks
Bill

A. Neumaier
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