Examples
The series
converges to 1/(1 − z) for |z| < 1. The Borel transform is
So the Borel sum is
which converges to 1/(1 − z) in the larger region Re(z) < 1, giving an analytic continuation of the original series.
The series
does not converge for any nonzero z.
The Borel transform is
for |t| < 1, and this can be analytically continued to all t ≥ 0. So the Borel sum is
(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.
Read more about this topic: Borel Summation
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