3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Your definition of Abel summability seems wrong. As written, every summable sequence $(s_n)$ would be Abel summable to $0$. $\endgroup$ Commented Feb 24, 2015 at 10:54
  • $\begingroup$ The definition is from the book of Jacob Korevaar: Tauberian Theory from Section 1: Equation (1.2). $\endgroup$ Commented Feb 24, 2015 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. $\endgroup$ Commented Feb 24, 2015 at 11:14
  • $\begingroup$ i edited the question. $\endgroup$ Commented Feb 24, 2015 at 18:32

1 Answer 1

Reset to default
3
+50
$\begingroup$

$$ \begin{matrix} \text{Let} & F(x) = \sum\limits_{n=0}^\infty \frac{s_n}{n+1} x^{n+1} &\text{so}& F'(x) = \sum\limits_{n=0}^\infty s_n x^n \\ \text{and} & G(x) = -\log(1-x) &\text{so}& G'(x)=\frac1{1-x} \\ \text{then} & -\dfrac1{\log (1-x)}\sum\limits_{n=0}^\infty \dfrac{s_n}{n+1}x^n=\dfrac{F(x)}{G(x)} &\text{and}& (1-x)\sum\limits_{n=0}^\infty s_nx^n=\dfrac{F'(x)}{G'(x)}. \\ \end{matrix} $$

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 $\lim F$.)

$\endgroup$
2
  • 2
    $\begingroup$ 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? $\endgroup$ Commented Feb 27, 2015 at 18:16
  • 3
    $\begingroup$ 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 $0$. $\endgroup$ Commented Feb 28, 2015 at 7:22

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.