How do I prove that the Borel summation of an arbitrary convergent series, \begin{equation} S = \sum_{k=0}^{\infty} a_k \end{equation} defined as \begin{equation} \int_{0}^{\infty} e^{-t} \sum_{k=0}^{\infty}a_k\frac{t^k}{k!}dt \end{equation} always converges to $S$? To be clear, I want to know how to prove it when the series is not necessarily absolutely convergent.
The Wikipedia article on Borel summation here justifies switching the order of the integral and the sum with "absolute convergence". Am I just missing something or was it only proving it for taylor series on the interior of an analytic disc?
Let the sequence $b=b_1,b_2,..$ satisfy $b_n \to 0$ and let $f_b(t)=\sum_{k=0}^{\infty}b_ke^{-t} \frac{t^k}{k!}, t \ge 0$ (which is trivially absolutely convergent for a fixed $t$).
We claim that $f_b(t) \to 0, t \to \infty$. This is a standard proof and I will append it at the end, but now I want to show how this implies the required Borel convergence.
Assume $\sum a_n \to A$ and as usual let $s_n=\sum_{ k\le n}a_k \to A$ its partial sums.
Also as is standard in the theory of Borel convergence we call $\alpha(t)=\sum_{k=0}^{\infty}a_k \frac{t^k}{k!}, \beta(t)=\sum_{k=0}^{\infty}s_k \frac{t^k}{k!}$ and we note that $\beta'(t)-\beta(t)=\alpha'(t)$
Applying the result above first with $b_n=s_n-A \to 0$ we get that $e^{-t}\beta(t)-A \to 0, t \to \infty$, while applying it with $b_n =a_n \to 0$ also, we get $e^{-t}\alpha(t) \to 0, t \to \infty$
A simple integration by parts shows that:
$e^{-t}\beta(t)-a_0=\int_0^{t}d(e^{-r}\beta(r))=\int_0^{t}e^{-r}\alpha'(r)dr=e^{-t}\alpha(t)-a_0+\int_0^{t}e^{-r}\alpha(r)dr$
Using the above and letting $t \to \infty$ we get that $\int_0^{t}e^{-r}\alpha(r)dr \to A, t \to \infty$ which is exactly
\begin{equation} \int_{0}^{\infty} e^{-t} \sum_{k=0}^{\infty}a_k\frac{t^k}{k!}dt =A \end{equation}
so we are done up to the simple fact regarding $f_b$ in the beginning
To prove that $f_b(t) \to 0, t \to \infty$ we pick $\epsilon >0$ and $N=N(\epsilon), |b_n| \le \epsilon, n\ge N$.
Then $|\sum_{k=N}^{\infty}b_ke^{-t} \frac{t^k}{k!}| \le \epsilon\sum_{k=N}^{\infty}e^{-t} \frac{t^k}{k!} \le \epsilon$ independent ot $t$ and then we can pick $t>T(\epsilon)$ s.t each of the missing terms in $f_b(t)$ so $b_ke^{-t} \frac{t^k}{k!}, k=0,..N-1$ is at most $\epsilon/N$ in absolute value since $e^{-t} \frac{t^k}{k!} \to 0, t \to \infty$ for each fixed $k$. Putting all together, we get $T(\epsilon)$ s.t $|f_b(t)| \le 2\epsilon, t \ge T(\epsilon)$ and we are done with this part too!