TITLE: Abel/Cesaro summable implies Borel summable?
QUESTION [6 upvotes]: Does Abel or Cesaro summable imply Borel summable for a series? In other words, for a sequence $(a_n)$ and its partial sums $(s_n)$, is it true that:
$$
\begin{split}
\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^{n-1} s_k &= A \implies \lim_{t \to \infty}e^{-t}\sum_{n=0}^{\infty}s_n\frac{t^n}{n!} = A\\
\lim_{x \to 1^-}\sum_{n=0}^{\infty}a_nx^n &= A \implies \lim_{t \to \infty}e^{-t}\sum_{n=0}^{\infty}s_n\frac{t^n}{n!} = A
\end{split}\;?
$$
Is there a proof of this if it is true?
If it isn't, then is there a sequence which is Abel/Cesaro summable but not Borel summable, and is Borel summability consistent with Abel/Cesaro summability?

REPLY [3 votes]: No, according to page 33 of "Borel's methods of summability" by B. Shaywer and B. watson, it follows that the domains of Cesaro (/Abel) summable series and (weak & strong) Borel summable series are not contained in each other. I don't have good examples for the concerned series. 
There do however exist Tauberian theorems stating the conditions on which a series summable by the one method implies it is summable by the other method.
On the series where both methods are assign a finite value, they do assign the same value hence they are consistent methods.