Is my proof of Abel's Theorem correct? Abels Theorem says that if $f(x)=\sum_{n=0}^{\infty} a_nx^n$ converges for $|x|<1$ and if $\sum_{n=0}^{\infty} a_n$ converges then $f$ is continuous at $x=1$. 
Proof.
Suppose $f(x)=\sum_{n=0}^{\infty} a_nx^n$ converges for $|x|<1$, and $\sum_{n=0}^{\infty} a_n$.
Subtract a constant from $f$ so that we may assume $\sum_{n=0}^{\infty} a_n=0$.
Let $A_n=\sum_{j=0}^{n} a_n$ 
We wish to show that $\lim_{x\to 1^{-}} |\sum_{n=0}^{\infty} a_nx^n|=0$
$\sum_{n=0}^{\infty} a_nx^n=a_0+ \sum_{n=1}^{\infty} a_nx^n= \sum_{n=1}^{\infty} (A_n-A_{n-1})x^n+a_0$
$=\sum_{n=1}^{\infty} A_nx^n - \sum_{n=1}^{\infty} A_{n-1}x^n +a_0 = (1-x)\sum_{n=0}^{\infty} A_nx^n$
Now, $|(1-x)\sum_{n=1}^{\infty} A_nx^n| = |(1-x)\sum_{n=0}^{k} A_nx^n + (1-x)\sum_{n=k+1}^{\infty} A_nx^n|$
For the first chunk, $|(1-x)\sum_{n=0}^{k} A_nx^n|$, since this summation is a finite quantity we choose $\delta_1$ such that $|x-1|<\delta_1 \implies |(1-x)\sum_{n=0}^{k} A_nx^n|<\epsilon/2$.
For the second chunk, $|(1-x)\sum_{n=k+1}^{\infty} A_nx^n| \leq |(1-x)Sup|A_n|\sum_{n=k+1}^{\infty} x^n| \leq |Sup|A_n|x^{k+1}|$ 
But $\lim_{n\to\infty} A_n = \lim_{n\to\infty} \sum_{j=0}^{n} a_n = 0$, so the second chunk can be made small as desired.
End Proof
I have been trying to understand this proof for quite a long time so any feedback is greatly appreciated. I am particularly concerned with the step that involved subtracting a constant from $f$ in order to suppose the summation converges to zero. 
 A: Given $\epsilon >0,$ take the least (or any) $k$ such that $\sup_{n>k}|A_n|<\epsilon /2.$ This is possible because $\lim_{n\to \infty}A_n=\sum_{m=0}^{\infty}a_m=0.$
Let $M_k=\max_{n\leq k}|A_n|.$ For $x\in (0,1)$ we have $|(1-x)\sum_{n=0}^kA_nx^n|\leq (1-x)(k+1)M_k.$
Take $\delta_k \in (0,1)$ such that $\delta_k(k+1)M_k<\epsilon /2.$ Then for all $x\in (1-\delta_k,1)$  we have $$|(1-x)\sum_{n\leq k}A_nx^n|<\epsilon /2$$ $$\text {and also }\quad |(1-x)\sum_{n>k}A_nx^n|\leq (1-x)\sum_{n>k}(\epsilon /2)x^n=$$ $$=x^{k+1}(\epsilon/2)<\epsilon /2.$$
So the theorem holds  in the first case, which is when $\sum_{n=0}^{\infty}a_n=0 .$ For all other cases observe that if $a^*_0=a_0-\sum_{n=0}^{\infty}a_n$ and $a^*_n=a_n$ for all $n>0,$ then the first case applies to $f^*(x)=\sum_{n=0}^{\infty} a^*_nx^n.$ That is, $f^*$ is continuous from below at $x=1.$ But $f(x)$ differs from $f^*(x)$ by a constant so $f$ is also continuous from below at $x=1.$
We can do  a direct proof for all cases at once. It's basically the same proof, but the intermediate formulas get a bit "messy" or confusing.
