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Let $0\leq x<1$ and $s_n$ be a sequence of partial sums of the series $\sum_{n=0}^{\infty}a_n$. It is called that the series $\sum_{n=0}^{\infty}a_n$ is $(A)$ or Abel summable to $s$ if $$\lim_{x\to1^-}(1-x)\sum_{n=0}^{\infty}s_nx^n=s,$$ and the series $\sum_{n=0}^{\infty}a_n$ is called $(L)$ summable to $s$ if $$\lim_{x\to1^-}\frac{-1}{\log(1-x)}\sum_{n=0}^{\infty}\frac{s_n}{n+1}x^{n+1}=s.$$

I need help to prove $(A)$ summability of the series $\sum_{n=0}^{\infty}a_n$ to $s$ implies $(L)$ summability of the series $\sum_{n=0}^{\infty}a_n$ to $s$. That is $(L)$ summability includes $(A)$ summability.

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  • $\begingroup$ Your definition of Abel summability seems wrong. As written, every summable sequence $(s_n)$ would be Abel summable to $0$. $\endgroup$ – Daniel Fischer♦ Feb 24 '15 at 10:54
  • $\begingroup$ The definition is from the book of Jacob Korevaar: Tauberian Theory from Section 1: Equation (1.2). $\endgroup$ – Raio Feb 24 '15 at 11:06
  • $\begingroup$ I can't see Section 1, but from examples 2.2 and 2.3, it seems that the sequence $(a_n)$ is Abel summable to $s$ if $$\lim_{x\to 1^-} (1-x)\sum_{n=0}^\infty s_n x^n = s,$$ where the $(s_n)$ are the partial sums, $$s_n = \sum_{k=0}^n a_k.$$ That looks more sensible. $\endgroup$ – Daniel Fischer♦ Feb 24 '15 at 11:14
  • $\begingroup$ i edited the question. $\endgroup$ – Raio Feb 24 '15 at 18:32
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$$ \begin{matrix} \text{Let} & F(x) = \sum\limits_{n=0}^\infty \frac{s_n}{n+1} x^{n+1} &\text{so}& F'(x) = \sum\limits_{n=0}^\infty s_n x^n \\ \text{and} & G(x) = -\log(1-x) &\text{so}& G'(x)=\frac1{1-x} \\ \text{then} & -\dfrac1{\log (1-x)}\sum\limits_{n=0}^\infty \dfrac{s_n}{n+1}x^n=\dfrac{F(x)}{G(x)} &\text{and}& (1-x)\sum\limits_{n=0}^\infty s_nx^n=\dfrac{F'(x)}{G'(x)}. \\ \end{matrix} $$

We want to prove that $\lim\limits_{x\to1-0} \dfrac{F'(x)}{G'(x)}=s$ implies $\lim\limits_{x\to1-0} \dfrac{F(x)}{G(x)}=s$.

As $\lim\limits_{x\to1-0}G(x)=\infty$, this is true by L'Hospital's rule.

(Most textbooks state L'Hospital's rule for limits of the forms $\frac00$ and $\frac{\pm\infty}{\pm\infty}$, but the case $\lim |G|=\infty$ no assumption is required on $\lim F$.)

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    $\begingroup$ I cant find any example for the converse case is not true. Namely, there exists a sequence which is summable (L) but not summable (A). Do you see any example for this case? $\endgroup$ – Raio Feb 27 '15 at 18:16
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    $\begingroup$ You can start with finding a suitable $F(x)$. For example $F(x)=\sin\frac1{1-x}$ is bounded so $\lim_{x\to1-0}\frac{F(x)}{-\ln(1-x)}=0$ but $F$ oscillates too rapidly: $(1-x)F'(x)=\frac{\cos\frac1{1-x}}{1-x}$ which has no limit. Now you can take the Taylor coefficients of $F'$ at $0$. $\endgroup$ – G. Kós Feb 28 '15 at 7:22

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