Proving absolute convergence by proving limit via power condition I was given the following premise:
If $\lim\limits_{x \to 1} \frac{|a_{n+1}|}{|a_n|} = r<1$, then $\sum a_n$ is absolutely convergent by the following: 
Show $\sum (r+\epsilon)^n < \infty$
I really don't know where to start to be honest, I've had "proof writer's block". I know how to define an $\ r+\epsilon \ $ neighborhood about r. I also know that $\lim\limits_{x \to 1} \frac{|a_{n+1}|}{|a_n|} = r<1$ in itself automatically implies that $a_n$ is convergent about the ratio test. I don't know how to incorporate the $(r+\epsilon)^n < \infty$ element of the proof. 

You have a typo in your definition, the quotient criteria is $\lim\limits_{n \to \infty} \frac{|a_{n+1}|}{|a_n|} = r<1$
In this case $a_n=(r + \varepsilon)^n$, so:
$$\lim\limits_{x \to 1} \frac{|a_{n+1}|}{|a_n|} = \frac{|r+\varepsilon|^{n+1}}{|r+\varepsilon|^n} = |r+\varepsilon|$$
And the series is absolutely convergent if $|r+\varepsilon|<1$. $r<1$, so $\exists \varepsilon >0$ such that $|r+\varepsilon|<1$ and you can choose this epsilon to get your convergent series.
