question about the ratio test proof part 1 If $\displaystyle\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n}}| = L < 1$, then the series converges (part 1 of ratio test).
Proof:
Since $L < 1$, $\exists r$ s.t. $0 \leq L < r < 1$. In particular,
$\displaystyle|\frac{a_{n+1}}{a_{n}}| < r \Rightarrow |a_{n+1}| < |a_{n}|r$.
For $n \geq N$, $|a_{N+2}| < |a_{N+1}|r < |a_{N}|r^2$.   ($\ast$)
The pattern will continue to $|a_{N+k}| < |a_{N}|r^k$.
I will just roughly describe the rest of the proof. $|a_{N}|r^k$ is a geometric series, hence converges and by the basic comparison test, $|a_{N+k}|$ will converge (absolutely).
Okay so the algebra in this proof makes sense to me, but there a couple of parts I still do not get. I do not understand why $r$ goes up a power each time at $\ast$. The line above that one, it was just $|a_{n}|r$. Why does it change? I also don't get why we replace $n$ with $N$ in the proof. I know $N$ is an integer but I still don't understand why we do it. I was always iffy on the whole $n > N$ idea. What is happening visually? Thanks!
 A: Suppose that $L>1$ and 
$$\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=L$$
Then by definition, this means that for all $\epsilon>0$ there exists a number $N$ such that whenever $n>N$ we have
$$L-\epsilon<\left|\frac{a_{n+1}}{a_{n}}\right|<L+\epsilon \tag 1$$
Note that here $\epsilon>0$ is any number given, whereas $N$, which depends on $\epsilon$, can be found (i.e., $N(\epsilon)$ exists).
Now since $(1)$ is true for all $\epsilon>0$, then it is true for $\epsilon'$ so small that $L+\epsilon' <r<1$ for some $r<1$.  
Now, given any $\epsilon>0$ with $\epsilon<\epsilon'$, there exists a number $N(\epsilon)$ such that whenever $n>N(\epsilon)$ 
$$\left|\frac{a_{n+1}}{a_{n}}\right|<r \tag 2$$

On the Questions Regarding $N$ and Increasing Powers of $r$
In particular, since $(2)$ is valid for all $n>N$, then starting with $n=N+1$, and proceeding with $n=N+2$, $n=N+3$, $\cdots n=N+m$, we certainly can write
$$\begin{align}
&\left|\frac{a_{N+2}}{a_{N+1}}\right|<r\implies |a_{N+2}|<r|a_{N+1}|\\\\
&\left|\frac{a_{N+3}}{a_{N+2}}\right|<r\implies |a_{N+3}|<r|a_{N+2}|<r^2|a_{N+1}|\\\\
&\left|\frac{a_{N+4}}{a_{N+3}}\right|<r\implies |a_{N+4}|<r|a_{N+3}|<r^2|a_{N+2}|<r^3|a_{N+1}|\\\\
&\vdots\\\\
&\left|\frac{a_{N+m}}{a_{N+m-1}}\right|<r\implies |a_{N+m}|<r^{m-1}|a_{N+1}|
\end{align}$$
Can you proceed from here?
