4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $z\to1^-$. $\endgroup$ – Calvin Khor Dec 14 '16 at 19:09
  • $\begingroup$ Is there anything about uniform convergence that would help to interchange the order of the limits? $\endgroup$ – AlexError Dec 14 '16 at 20:08
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.