How to prove this Tauberian theorem How to prove this exercise on tauberian theorem from Zorich:

Tauber's original theorem relates to Abel summation of series and consists
of the following.


Suppose the series $\sum\limits_{n=1}^\infty a_nx^n$ converges for $0 < x < 1$ and $\lim\limits_{x\to1-0}\sum\limits_{n=1}^{\infty}a_nx^n=A$. If $\lim\limits_{n\to \infty} \frac{a_1+2a_2+\cdots+na_n}{n}=0$, then the series $\sum\limits_{n=1}^\infty a_n$ converges to $A$ in the ordinary sence.

I can prove a weaker form of the theorem when
replace "If $\lim_{n\to \infty} \frac{a_1+2a_2+\cdots+na_n}{n}=0$" with "If $a_n=o(\frac1n)$" .
But for the exercice, I don't know if the following solution works:
(1).By Cauchy's criterion$\lim\limits_{x\to1-0}\sum\limits_{n=1}^{\infty}a_nx^n\Rightarrow \exists N$ s.t. $\sum\limits_{n=1}^{\infty}a_n[(1-\frac1N)^n-(1-\frac1{2N})^n]\leq\epsilon$
. Hence  $\sum\limits_{n=1}^{\infty}\frac{a_n(n+1)}{2N}(1-\frac1N)^{n-1}\leq \epsilon$
(2).$|\sum\limits_{n=1}^Na_n-\sum\limits_{n=1}^{\infty}a_n(1-\frac1N)^n|\leq|\sum\limits_{n=1}^{N}\frac{na_n}N|+|\sum\limits_{n=N+1}^{\infty}|a_n|(1-\frac1N)^n|$
By the assumption, we have $|\sum\limits_{n=1}^{N}\frac{na_n}N|\leq\epsilon$ for big enough $N$, but how should I estimate the second item $|\sum\limits_{n=N+1}^{\infty}a_n(1-\frac1N)^n|$?
I guess we might need to use (1). to control it, or use the property of $\sum a_nx^n$ convergent absolutely in (0,1), but I failed to work it out.
I know this is a theorem proved by Tauber but I can't find any proof (in English) about it. Feel free to give any hints or solutions, or links about the proofs of this theorem.
 A: We can in fact use the proved theorem with condition $a_n=o(\frac1n)$.
Let $b_n=\sum\limits_{k=1}^nka_k$, hence $\lim\limits_{n\to\infty}\frac{b_n}{n}=0$.[we let $\frac{b_0}*=0$]
\begin{align}\sum\limits_{n=1}^{\infty}a_nx^n=&\sum_{n=1}^{\infty}\frac{b_n-b_{n-1}}{n}x^n=\sum_{n=1}^{\infty}(\frac{b_n}{n}-\frac{b_{n-1}}{n-1}+\frac{b_{n-1}}{n-1}-\frac{b_{n-1}}{n})x^n\\=&(1-x)\sum_{n=1}^\infty\frac{b_n}{n}x^n+\sum_{n=1}^{\infty}\frac{b_n}{n(n+1)}x^{n+1}\tag1\end{align}
besides, for big enough N, we have $n>N(\frac{b_n}n\leq\epsilon)$:\begin{align}(1-x)\sum_{n=1}^\infty\frac{b_n}{n}x^n=&(1-x)\sum_{n=1}^N\frac{b_n}nx^n+(1-x)\sum_{n=N+1}^\infty\frac{b_n}{n}x^n\\\leq&(1-x)\sum_{n=1}^N\frac{b_n}nx^n+(1-x)\sum_{n=N+1}^{\infty}\epsilon x^n\\\leq&(1-x)\sum_{n=1}^N\frac{b_n}nx^n+\epsilon\end{align}
So:
$$\lim_{x\to1-0}(1-x)\sum_{n=1}^\infty\frac{b_n}{n}x^n=0$$
Note that $\lim\limits_{n\to\infty}n\cdot\frac{b_n}{n(n+1)}=0$, and let $x\to1-0$ in the above equation (1), and use the "weaker form" proved:
$$A=0+\sum_{n=1}^{\infty}\frac{b_n}{n(n+1)}$$
But $$\sum_{n=1}^{M}\frac{b_n}{n(n+1)}=\sum_{n=1}^M\frac{b_n}{n}-\sum_{n=1}^{M+1}\frac{b_{n-1}}{n}=-\frac{b_{M}}{M+1}+\sum_{n=1}^{M}a_n$$
Hence $\sum_{n=1}^{\infty}a_n=A$.
