Can Borel Resummation Solve Complex Series Summation?

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SUMMARY

Borel resummation provides a method to sum divergent series by transforming them into convergent forms. Specifically, if a series is represented as \( f(x) = \sum_{n=0}^{\infty} \frac{a(n)}{n!}x^{n} \), the sum can be obtained through the integral \( S = \int_{0}^{\infty} dte^{-t}f(t) \), contingent on the convergence of the integral. However, complications arise when the coefficients \( a(n) \) are complex or when \( f(t) \) grows faster than any positive exponential, leading to potential issues with convergence and necessitating numerical integration methods.

PREREQUISITES
  • Understanding of divergent series and their properties
  • Familiarity with Borel resummation techniques
  • Knowledge of integral calculus, particularly improper integrals
  • Basic concepts of asymptotic analysis and growth rates of functions
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lokofer
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Borel "resummation"...

Let be a divergent series:

[tex]\sum _{n=0}^{\infty} a(n)[/tex] (1)

then if you "had" that [tex]f(x)= \sum _{n=0}^{\infty} \frac{a(n)}{n!}x^{n}[/tex]

You could obtain the "sum" of the series (1) as [tex]S= \int_{0}^{\infty}dte^{-t}f(t)[/tex] in case the integral converges...

- Yes that's "beatiful" the problem is ..what happens if the coefficients a(n) are complicate?..then how can you obtain the sum of the series?...

- By the way i think that Borel resummation can be applied if [tex]f(t)=O(e^{Mt})[/tex] M>0, but what happens if f(t) grows faster than any positive exponential?.. :cry:
 
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If ##f(x)## is "complicated", then there will be only a numerical integration possible. And if ##f(x)## grows faster than exponential, then the integral likely does not converge.
 

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