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.
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 $\sum_{n=1}^{\infty}a_n=A$.