Hey, I have a question about the proof of the ratio test. I'm using the Stewart Calculus 6thed book, which gives the following for the proof:

Choose a number r such that L<r<1.

Since \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right | =L and L<r

\left | \frac{a_{n+1}}{a_n} \right | will eventually be less than r if there exists an int. N

\left | \frac{a_{n+1}}{a_n} \right | < r whenever n\ge N

which is also:

|a_{n+1}|<|a_{n}|r whenever n \ge N

then we do the following:

|a_{N+1}|<|a_{N}|r
|a_{N+2}|<|a_{N+1}|r<|a_{N}|r^2
|a_{N+3}|<|a_{N+2}|r<|a_{N}|r^3

Note: I don't exactly get what is going on here, but it makes a little sense when you take |a_{n}|<Ar^n \Rightarrow |a_{n}|r<Ar^{n+1}, I got that part from another book.

so: |a_{N+k}|<|a_{N}|r^k

now we take the series: \sum_{k=1}^{\infty}|a_{N}|r^{k}=|a_{N}|r+|a_{N}|r^  {2}+|a_{N}|r^{3}+...

Which is convergent because it is a geometric series where 0<r<1

Using the comparison test, we see that \sum_{n=N+1}^{\infty}|a_{n}|=\sum_{k=1}^{\infty}|a  _{N+k}|=|a_{N+1}|+|a_{N+2}|+|a_{N+3}|+... is also convergent. So the series \sum_{n=1}^{\infty}|a_n| is convergent, which means that \sum_{n=1}^{\infty}a_n is absolutely convergent, which also means it is convergent.

Now, if \left |\frac{a_{n+1}}{a_n}\right | \rightarrow L > 1 or \left |\frac{a_{n+1}}{a_n}\right | \rightarrow \infty and there is an int. N such that:

\left |\frac{a_{n+1}}{a_n}\right | > 1 whenever n\ge N

This means that |a_{n+1}|>|a_n| whenever n\ge N. So \lim_{n \to \infty}a_n \ne 0

and if \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right | =1, there it is inconclusive because it may diverge or converge, the book gives a few examples why.

Now, I think I get how they come to the conclusion, why it would work using the comparison test to test the sequences, and figuring out that the sequence in the limit is less than the geometric sequence; but I'm confused as to how they go from comparing a limit with a series, i.e. how they determined that the limit \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right | =L could prove that \sum_{n=1}^{\infty}|a_n| is convergent because \sum_{n=1}^{\infty}ar^{n-1} is convergent. It's like comparing apples and oranges to me.

Considering this is a known proof, It's my ignorance at fault, but I'm hoping someone can shed some light on this.

Thanks!