Proof of the Ratio Test

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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:

  1. if L < 1, then the series converges
  2. if L > 1, then the series diverges
  3. if L = 1, then the test is inconclusive

Proof

Our proof will be in two parts:

  1. Proof of 1 (if L < 1, then the series converges)
  2. 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 |an+1/an| 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,

Then the series

is convergent because it is a geometric series with a common ratio r, such that 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)

Our aim here is to show that the series must be divergent using the Divergence Test.

If

then the ratio

will eventually be greater than 1; that is, there exists an integer N such that

This means |an+1| > |an| whenever nN, and so by the definition of limit

Therefore, the given series diverges by the Divergence Test.


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