# 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:

$\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^{n-1} s_k = A \Longrightarrow \lim_{t \to \infty}e^{-t}\sum_{n=0}^{\infty}s_n\frac{t^n}{n!} = A$ $\lim_{x \to 1^-}\sum_{n=0}^{\infty}a_nx^n = A \Longrightarrow \lim_{t \to \infty}e^{-t}\sum_{n=0}^{\infty}s_n\frac{t^n}{n!} = A$.

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 $z\to1^-$. – Calvin Khor Dec 14 '16 at 19:09
• Is there anything about uniform convergence that would help to interchange the order of the limits? – AlexError Dec 14 '16 at 20:08