Proving Convergence and Absolute Convergence of Power Series How do you prove the following claim?
If a power series $\sum_{n=0}^{\infty} a_n (x-a)^n$ converges at some point $b ≠ a$, then this power series converges absolutely at every point closer to $a$ than $b$ is.
Here's what I tried so far. Does this proof make sense? What would you change to hone it further?  
Proof: 
Suppose that $\sum_{n=0}^{\infty} a_n (x-a)^n$ converges at $b ≠ a$. 
Then, the $ \lim\limits_{x\to \infty} |a_n (b-a)^n|=0$
For $ɛ=1>0$, $∃N$ such that, if $n>N$, then $-ɛ<|a_n (b-a)^n|<ɛ$ 
That is to say, $|a_n (b-a)^n|<1 (*)$
Take any $x$, such that $|x-a| < |b-a| $ Implicitly, $\frac{| (x-a)|}{|(b-a)|}<1 (**)$
For $n>N$, notice that $|a_n (x-a)^n| = |a_n (x-a)^n| × \frac{|a_n (b-a)^n|}{|a_n (b-a)^n|}$
Rearranging, we get: 
$|a_n (b-a)^n| × \frac{| (x-a)^n|}{|(b-a)^n|} < 1× \frac{| (x-a)^n|}{|(b-a)^n|} $  because of $(*)$
$\sum_{n=0}^{\infty} |a_n (x-a)^n| =\sum_{n=0}^{N} |a_n (x-a)^n| +\sum_{n=N+1}^{\infty} |a_n (x-a)^n|
< \sum_{n=0}^{N} |a_n (x-a)^n| +\sum_{n=N+1}^{\infty} \frac{| (x-a)^n|}{|(b-a)^n|}$
But, $\sum_{n=N+1}^{\infty} \frac{| (x-a)^n|}{|(b-a)^n|}$ is a geometric series, with ratio $<1$ (from **) 
Therefore, $\sum_{n=N+1}^{\infty} \frac{| (x-a)^n|}{|(b-a)^n|}$ converges. 
Hence, $\sum_{n=0}^{\infty} |a_n (x-a)^n| $ also converges (by the comparison test). 
Consequently, by definition, $\sum_{n=0}^{\infty} a_n (x-a)^n$ converges absolutely.  

There's something a bit strange about your proof.  Look at the line after "Rearranging, we get:"..  That inequality you say is because of (*) but if you cancel $\frac{|(x-a)^n|}{|(b-a)^n|}$ it's simply equivalent to $|a_n(b-a)^n|<1$.  So you don't need to go to all that work to get to this inequality.
Also, how do you get $\sum_{n=N+1}^{\infty} |a_n (x-a)^n|
< \sum_{n=N+1}^{\infty} \frac{| (x-a)^n|}{|(b-a)^n|}$
I don't think that follows from what you showed pervious to that line.
