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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?

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  • $\begingroup$ 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, 2016 at 19:09
  • $\begingroup$ Is there anything about uniform convergence that would help to interchange the order of the limits? $\endgroup$
    – AlexError
    Dec 14, 2016 at 20:08

1 Answer 1

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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.

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