Science:Infinite Series Module/Appendices/Proof of The Ratio Test/Proof of The Ratio Test
We can now provide the proof of the ratio test. Recall the ratio test:
The Ratio Test 

To apply the ratio test to a given infinite series
we evaluate the limit
There are three possibilities:

Proof
Our proof will be in two parts:
 Proof of 1 (if L < 1, then the series converges)
 Proof of 2 (if L > 1, then the series diverges)
Proof of 1 (if L < 1, then the series converges)
Our aim here is to compare the given series
with a convergent geometric series (we will be using a comparison test).
In this first case, L is less than 1, so we may choose any number r such that L < r < 1. Since
the ratio a_{n+1}/a_{n} will eventually be less than r. In other words, there exists an integer N such that
This follows from the formal definition of limit, which not all calculus students have covered. Those students who have not covered a formal definition of limit may wish to consult a calculus textbook on this (any of the textbooks listed on the Recommended Resources page cover it).
We can rearranged our expression to
If we let equal N, N + 1, N + 2 in the previous equation we obtain
and, in general,
Now the series
is convergent because it is a geometric series whose common ratio is known to satisfy 0 < r < 1. By the Comparison Test, the series
is convergent, and so our given series
is also convergent (adding a finite number of finite terms to a convergent series will create another convergent series).
Therefore, our series is absolutely convergent (and therefore convergent).
Proof of 2 (if L > 1, then the series diverges)
Here the Divergence Test implies that the given series diverges. Indeed, if
then the ratio
will eventually be greater than 1; that is, there exists an integer N such that
For this same chaining these inequalities together shows that for all . This clearly implies that the sequence cannot converge to 0. Therefore, the given series diverges by the Divergence Test.