Borel summation

Remarks

Example: for $a_k = z^k$ with a complex number $z$, $\mathcal{B}(t) = e^{zt}$, and the integral converges exactly when $\operatorname{Re} z < 1$, with value $1/(1 - z)$. So the geometric series is Borel summable to $1/(1-z)$ on the half-plane $\operatorname{Re} z < 1$, also where it diverges, such as at $z = -1$, where the sum is $1/2$. Non-example: for $a_k = (k!)^2$, the Borel transform $\sum_k k! \, t^k$ has radius of convergence $0$, so clause 1 fails. The method is regular: a convergent series is Borel summable, and its Borel sum is its ordinary sum. For a power series $\sum_k a_k z^k$, the Borel sum is $\int_0^\infty e^{-t} \mathcal{B}(zt) \, dt$, which is a Laplace transform of $\mathcal{B}$ up to a change of variable. A weaker variant, the weak Borel method, uses the limit of $e^{-t} \sum_{n \ge 0} s_n t^n/n!$ as $t \to \infty$, where $s_n$ are the partial sums.