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 such that .
Since and
will eventually be less than if there exists an int.
whenever
which is also:
whenever
then we do the following:
Note: I don't exactly get what is going on here, but it makes a little sense when you take , I got that part from another book.
so:
now we take the series:
Which is convergent because it is a geometric series where
Using the comparison test, we see that is also convergent. So the series is convergent, which means that is absolutely convergent, which also means it is convergent.
Now, if or and there is an int. N such that:
whenever
This means that whenever . So
and if , 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 could prove that is convergent because 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!