I have been trying to prove the following assertion:
"Let $a_n\in\mathbb{C}$ for $n\in\{0,1,2,\ldots\}$ and $s_n=\sum\limits_{k=0}^na_n$. Define formally two series $$ a(x)=\sum\limits_{n=0}^{\infty}\frac{a_nx^n}{n!} \quad\text{and}\quad s(x)=\sum\limits_{n=0}^{\infty}\frac{s_nx^n}{n!}.$$ Then if one series converges for all $x$ (i.e. has radius of convergence infinity), so does the other series."
I have so far been successful in proving one direction:
Assume the second series is convergent for all $x$, so the function $s(x)=\sum\limits_{n=0}^{\infty}\frac{s_nx^n}{n!}$ is well-defined. Then since this is a convergent power series, we can differentiate term by term to get $$s^{\prime}(x)=\sum\limits_{n=0}^{\infty}\frac{s_{n+1}x^n}{n!}, $$ which is also convergent for all $x$. Then the series defined as $$ s^{\prime}(x)-s(x) = \sum\limits_{n=0}^\infty\frac{a_{n+1}x^n}{n!} $$ must converge as well for all $x$, and for the same reason does the series resulting from integrating this series term by term, that is $$ \sum\limits_{n=1}^{\infty}\frac{a_nx^n}{n!} $$ Adding a constant $a_0$ doesn't influence the convergence and results in the series $a(x)$, hence we have proved convergence of this series.
As to the other direction, a similar argument doesn't work, since a sum of two divergent series can be a convergent one. I have tried expressing the Cauchy difference for the second series as $$ \sum\limits_{k=m}^n\frac{s_nx^n}{n!} = \sum\limits_{k=m}^{n}\sum\limits_{i=0}^ka^i\frac{x^k}{k!} $$ and changing the order of summation to get $$ \sum\limits_{i=0}^ma_i\sum\limits_{k=m}^n\frac{x^k}{k!}+\sum\limits_{i=m+1}^na_i\sum\limits_{k=i}^n\frac{x^k}{k!} $$ but even with this I could not conclude convergence (I attempted to construct the first sum in this expression as follows: $$ \sum\limits_{i=0}^m\frac{a_ix^i}{i!}\sum\limits_{k=m}^n\frac{x^{k-i}i!}{k!}+\sum\limits_{i=m+1}^n\frac{a_ix^i}{i!}\sum\limits_{k=i}^n\frac{x^{k-i}i!}{k!} $$ but could not conclude boundedness/convergence of the series $\sum\limits_{k=i}^n\frac{x^{k-i}i!}{k!}$, so it did not help much).
This assertion is from Hardy - Divergent Series (1949), section 8.5 in the proof that weak Borel transform and integral Borel transform are almost equivalent. The whole proof depends on this (and a few more after that), but Hardy does not prove this assertion, nor refers to any proof.
Do you have any ideas how to solve this?