Inclusion relation between two summability methods 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.

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