Abel's limit theorem using summation by parts I have the following problem:

I am trying to see how to approach part (iii) Abel's theorem.
I took the suggestion of @Halbaroth below. But, I don't get very far.
It looks like a simple substitution will give me $F(z_1)$, but it seems too straightforward and I did not do any translations or rotations to reduce the case considered in (ii). In (ii) $lim_{r \to 1−}$ in this case $lim_{r \to R−}$, how can I apply (ii) directly? I think I am still missing something - and its not a simple substitution. Basically by direct substitution, I have $lim_{r \to R^-} \sum_{0}^{\infty} F(z_0+re^{i \theta})=lim_{r \to R^-} \sum_{0}^{\infty} a_n(z_0 +re^{i \theta}-z_0)^n = F(z_1)=F(z_0+Re{i \theta}) = \sum_{0}^{\infty} a_n(z_0+Re^{i \theta} - z_0)^n$.
I don't think this is right.
 A: Assume $z_1 = z_0 + Re^{i\theta_0}$ and let $S = \sum_{n=0}^{\infty} a_n (z_1-z_0)^n$. Set $G(z) = F(z_0 + Re^{i\theta_0} z) - S$. Then $G$ is defined on the unit disk and its Taylor expansion is
$$ G(z) = a_0 - S + \sum_{n=1}^{\infty} a_n R^n e^{in\theta_0} z^n. $$
Let $(b_n)$ be the coefficients of the above series. It is clear that $\sum_{n=0}^{\infty} b_n = 0$. According to the second question and our substitution, we have
$$ \lim_{z \to z_1} F(z) - S = \lim_{z \to 1^-} G(z) = 0. $$
A: I believe that this is a pretty standard proof of the real version of Abel's Theorem

Summation by Parts
Let
$$
A_n=\sum_{k=0}^na_k\tag1
$$
Then
$$
\begin{align}
\sum_{k=0}^na_kr^k
&=A_0+\sum_{k=1}^n(A_k-A_{k-1})r^k\tag2\\
&=A_0(1-r)+A_nr^n+\sum_{k=1}^{n-1}A_kr^k(1-r)\tag3\\
&=A_nr^n+(1-r)\sum_{k=0}^{n-1}A_kr^k\tag4
\end{align}
$$
Assuming $\lim\limits_{n\to\infty}A_n=A$, we can set $M=\sup\limits_{n\ge0}|A_n-A|$. For any $\epsilon\gt0$, we can choose $n_\epsilon$ so that if $k\ge n_\epsilon$, then $|A_k-A|\le\epsilon$. Therefore, for $0\le r\lt1$,
$$
\begin{align}
\left|\,\sum_{k=0}^\infty a_kr^k-A\,\right|
&=(1-r)\left|\,\sum_{k=0}^\infty(A_k-A)r^k\,\right|\tag5\\
&\le(1-r)\sum_{k=0}^{n_\epsilon-1}|A_k-A|r^k+(1-r)\sum_{k=n_\epsilon}^\infty|A_k-A|r^k\tag6\\[6pt]
&\le\left(1-r^{n_\epsilon}\right)\!M+r^{n_\epsilon}\epsilon\tag7
\end{align}
$$
Explanation:
$(5)$: take the limit of $(4)$ as $n\to\infty$ and apply the sum of a geometric series
$(6)$: triangle inequality
$(7)$: apply the bounds on $|A_k-A|$

Taking the Limit in $\boldsymbol{r}$
Taking the limsup of $(7)$ as $r\to1^-$, we get
$$
\limsup_{r\to1^-}\left|\,\sum_{k=0}^\infty a_kr^k-A\,\right|\le\epsilon\tag8
$$
Since $(8)$ is true for any $\epsilon\gt0$, we have
$$
\begin{align}
\lim_{r\to1^-}\sum_{k=0}^\infty a_kr^k
&=A\tag9\\
&=\sum_{k=0}^\infty a_k\tag{10}
\end{align}
$$

Extending to $\boldsymbol{\mathbb{C}}$
Apply $(10)$, setting $a_k$ to the real and imaginary parts of $c_k(z_1-z_0)^k$, and we get that if
$$
\sum_{k=0}^\infty c_k(z_1-z_0)^k\tag{11}
$$
converges, then
$$
\lim_{r\to1^-}\sum_{k=0}^\infty c_k(z_1-z_0)^kr^k=\sum_{k=0}^\infty c_k(z_1-z_0)^k\tag{12}
$$
