# Abel/Cesaro summable implies Borel summable?

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?

• For any $0<z<1$, $$\lim_{t→∞}e^{-t} \sum_{n=0}^∞ \frac{t^n}{n!}\sum_{k=0}^na_kz^k = \sum_{k=0}^\infty a_k \lim_{t→∞}\left( e^{-t}\sum_{n=k}^∞ \frac{t^n}{n!}\right) z^k = ∑_{k=0}^∞ a_k z^k$$ but I'm not sure how to interchange the limit of $t→\infty$ and $z\to1^-$. Commented Dec 14, 2016 at 19:09
• Is there anything about uniform convergence that would help to interchange the order of the limits? Commented Dec 14, 2016 at 20:08