Showing $\lim_{r \to 1^-} \sum a_nr^n=A$ using Abel's theorem Theorem: Let $\sum a_n(z-a)^n$ have radius of convergence $1$ and suppose that $\sum a_n$ converges to $A$. Prove that $\lim_{r \to 1^-} \sum a_nr^n=A$
I found this answer Abel's Theorem, alternate proof
But I'm not sure I understand it. Why
$$S_N=\sum_{n=1}^N r^n a_n = r^N \sum_{n=1}^Na_n+ (1-r)\sum_{n=1}^{N-1}r^n\sum_{j=1}^{n}a_j$$
and then how he got from the above equation to this
$$a_N\frac{1-r^N}{1-r}+\sum_{n=1}^{N-1}(a_n-a_{n+1})\frac{1-r^{n+1}}{1-r}.
$$
the rest, he lost me completely. 
Here is his proof, just in case you missed the link
Prove by partial summation:
$$
S_N=\sum_{n=1}^N r^n a_n = r^N \sum_{n=1}^Na_n+ (1-r)\sum_{n=1}^{N-1}r^n\sum_{j=1}^{n}a_j\\ 
=a_N\frac{1-r^N}{1-r}+\sum_{n=1}^{N-1}(a_n-a_{n+1})\frac{1-r^{n+1}}{1-r}.
$$
Taking $N\to \infty$ first (as we have to...)
$$
S=\sum_{n=1}^\infty = (1-r) \sum_{n=1}^\infty r^n \sum_{j=1}^na_j\\
=\frac{1}{1-r}\sum_{n=1}^\infty (a_n-a_{n+1})(1-r^{n+1}).
$$
I really want to understand this to prepare for the final exam, so any help will be greatly appreciated.
 A: For this problem, we need to use Abel's lemma:

Let $S_n = \sum_{i=m}^na_i$ be the partial sum. Then
  $$
\sum_{i=m}^na_ib_i=\sum_{i=m}^{n-1}S_i(b_i-b_{i+1})+S_nb_n
$$

Let $\epsilon > 0$ be given. Let $S_N = \sum_1^Na_n$ be the partial sum. Then there exists $k,m>N$ such that $\Bigl\lvert\sum_{j=k}^ma_j\Bigr\rvert < \epsilon$. By Abel's lemma, we can write $\sum a_nr^n$ as
\begin{align}
\sum_{j=k}^na_jr^j &= \sum_{j=k}^{n-1}\Bigl[\sum_{j=k}^ma_j(r^j-r^{j+1})\Bigr]+r^n\sum_{j=k}^na_j\\
&= (1-r)\sum_{j=k}^{n-1}\sum_{j=k}^ma_jr^j+r^n\sum_{j=k}^na_j
\end{align}
For $r\in(0,1)$, the geometric series $\sum_1^{N-1} a_jr^j=\frac{1-r^N}{1-r}$ converges. We have
\begin{align}
\Bigl\lvert\sum_{j=k}^na_jr^j\Bigr\rvert & < (1-r)\sum_{j=k}^{n-1}\Bigl\lvert\sum_{j=k}^ma_j\Bigr\rvert r^j + r^n\Bigl\lvert\sum_{j=k}^na_j\Bigr\rvert\\
&< (1-r)\epsilon\sum_{j=k}^{n-1}r^j+r^n\epsilon\\
&< (1-r)\frac{1-r^n}{1-r}\epsilon+r^n\epsilon\\
&= \epsilon
\end{align}
Let $n\to\infty$. Then we have $\Bigl\lvert\sum_{j=k}^{\infty}a_jr^j\Bigr\rvert < \epsilon$.
Let $A=\sum_0^{\infty}a_n$. Then if your limit converges, 
$$
\Bigl\lvert\sum_{j=0}^{\infty}a_jr^j-A\Bigr\rvert < \epsilon
$$
Let's determine if it converges.
\begin{align}
\Bigl\lvert\sum_{j=0}^{\infty}a_jr^j-A\Bigr\rvert &\leq
\overbrace{\Bigl\lvert\sum\sum a_jr^j(1-r)\Bigr\rvert + \Bigl\lvert r^n\sum a_j\Bigr\rvert}^{\text{by Abel's lemma}} + \overbrace{\Bigl\lvert\sum a_k\Bigr\rvert}^{=A}\\
& < \sum\lvert a_j\rvert(1-r)\frac{1-r^n}{1-r} + r^n\epsilon + \epsilon\\
& < \sum\lvert a_j\rvert(1-r^n) +r^n\epsilon + \epsilon\tag{1}
\end{align}
Taking the limit as $r\to 1^-$ of equation $(1)$ yields 
$$
\sum\lvert a_n\rvert(1-r^n) = 0 < \epsilon
$$
Let $\epsilon = \epsilon/3$ and as $r\to 1^-$, we have
$$
\Bigl\lvert\sum_{j=0}^{\infty}a_jr^j-A\Bigr\rvert < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon
$$
Therefore, the limit converges to $A$.
A: Hint : Write $\sum_{n=1}^N r^n a_n=\sum _{n=1}^Nr^n(D_{n+1}-D_{n})$ where $D_n=\sum _{k=1}^{n-1}a_k$ and use the principle of Abel's transform. For the second equality write $\sum_{n=1}^N r^n a_n=\sum_{n=1}^N (R_{n+1}-R_{n}) a_n$ where $R_n=\sum _{k=0}^{n-1}r^k=\frac{1-r^{n}}{1-r}$.
