Why is the borel sum analytic I am currently reading in a book about Borel sums as a method of analytic continuation of power series. So given a power series $\sum_{n=0}^{\infty}a_nz^n$ the borel sum is defined as $\int_0^\infty e^{-u} \sum_{n=0}^{\infty}\frac{a_nz^nu^n}{n!}du$. But the book does not give a justification why the function defined by the borel sums is actually analytic in the region where the integral converges. Is there an easy way to see this?
 A: Let $R$ be the radius if convergence and let $|z| <R$. Let $|z| <R_1<R_2<R$. Since $\sum a_n R_2^{n}$ is convergent, the sequence $|a_n|R_2^{n}$ tends to $0$. In particular there exists a constant $C$ such that $|a_n|R_2^{n}\leq C$. Now $|a_nz^{n}| \leq C (\frac {R_1} {R_2})^{n}$. Hence $\int_0^{\infty}e^{-u} |\sum \frac {a_nz^{n}u^{n}} {n!}|du$ is dominated by $C\int_0^{\infty}e^{-u}e^{\frac {R_1} {R_2}u}du<\infty$.
A: As I did some research into the topic in the last few days, I just wanted to improve the given answer a little bit for future reference: 
Given a power series $\sum_{n=0}^\infty a_n z^n$ with positive radius of convergence, the region where the borel sum is analytic can be described very precisely. It is the so-called borel polygon which is constructed in the following way: For any ray starting at the origin (or more general the expansion point of the power series) which has a singularity on it, take the nearest singularity (with respect to the origin) and consider the line crossing this singularity and being perpendicular to the ray. This line divides the plane into two areas and we take the area containing the origin.
If we do that for all rays and intersect all these areas, we get the borel polygon. Note that despite of its name it does not have to be a polygon.
Some examples for borel polygons
Outside the borel polygon the borel sum diverges and on the boundary of the borel polygon convergence and divergence is possible.
Proofs can be found in "Tauberian Theory: A Century of Developments" by Jacob Korevaar (283ff.) and "Divergent Series" by G. H. Hardy (178ff.). I found the latter freely available here.
But to answer the original question: No there is no easy way to see it, one has to develop some theory.
