The Ratio Test tests a series for convergence or divergence by considering the limit of successive terms. It is an important test: For example, it's frequently used in finding the interval of convergence of power series.
Theorem. ( Ratio Test)
Let be a series with
positive terms. Let
(a) If , the series converges.
(b) If , the series diverges.
(c) If , the test fails.
The reason the test works is that, in the limit, the series looks like a geometric series with ratio L. You can see that the inequalities for L resemble the inequalities that tell when a geometric series with (positive) ratio r converges or diverges.
The proof will use the following result on sequences: An increasing sequence that is bounded above has a limit.
Proof. I'll consider the case by way of example. Choose a positive number
so
. For n
sufficiently large,
These inequalities give
Adding the inequalities yields
The right side is a convergent geometric series. The inequality shows
that its sum is an upper bound for the partial sums of the series on
the left. The partial sums form an increasing sequence that is
bounded above, so they have a limit --- that is, the series on the
left converges. Hence, the original series converges, since
it's just
This is a finite number (
) plus the series
,
which I know converges.
A similar argument works if .
When do you use the Ratio Test? Ratios are fractions, and they tend
to simplify nicely if the top and bottom contain products or
powers. For example, if the term of the
series contains factorials, you ought to give the Ratio Test
serious consideration.
Example. Does converge
or diverge?
I'll approach this example as if it didn't appear in a discussion of the Ratio Test. What do you do?
The Zero Limit Test is easy to apply. However,
Hence, the Zero Limit Test fails.
The series is not geometric, and it's not a p-series.
The Integral Test is inapplicable. What would mean as a continuous function? How would you
integrate it?
It's possible to apply a comparison test; do you see how?
The Ratio Test is probably the easiest way to show that this series
converges. One indication that the Ratio Test is worth trying is that
is a product. The Ratio Test works well with
products and powers, because cancellation may occur when you form
.
Form the ratio of successive terms:
Take the limit as :
The limit is less than 1. The series converges, by the Ratio
Test.
Example. Does converge or diverge?
First, note that is the product of the
numbers from 1 to
:
For example, if ,
, and
I'll apply the Ratio Test. Note that if I replace n with in
, I get
. So I have
I'll stop for a second and show the details of the next simplification:
Thus, my limit is
The limiting ratio is less than 1, so the series converges by the
Ratio Test.
Example. Does
converge or diverge?
The Zero Limit Test fails.
The series is not geometric, and it's not a p-series. I don't think
you'd want to integrate ! And it's
not clear how to do this using a comparison.
Form the ratio of successive terms:
Take the limit as :
Since , the series converges, by the
Ratio Test.
Example. Does the series converge or diverge?
What happens if I try to use the Ratio Test? The limiting ratio is
The Ratio Test fails.
In general, the Ratio Test will fail if the general term is a rational function.
In this case, limit comparison is a better choice. Since , I'll compare the given series to
:
The limit is a finite positive number.
converges, since it's a p-series with
. Hence, the original series converges by Limit
Comparison.
The Root Test is similar to the Ratio Test.
Instead of taking the limit of successive quotients of terms, you
take the limit of the root of the
term.
Theorem. ( Root Test)
Let be a series with
positive terms. Let
(a) If , the series converges.
(b) If , the series diverges.
(c) If , the test fails.
Informally, the test works for the same reason that the Ratio Test
works --- namely, when k is large, , so
. This says that the series is
approximately geometric for large k, so it converges if the ratio L
is less than 1 and diverges if the ratio L is greater than 1.
Proof. I'll sketch a proof in the case where
. So suppose
I want to show that the series diverges. Pick a number r such that
. Then
means that I can find a number n
so that if
I have
, or
. Then
That is, the partial sums of the series are each greater
than the corresponding partial sums of
. But the series
is a divergent
geometric series, since
. Therefore, its partial
sums go to
, and hence the partial sums of
go to
as well. So
diverges, which means the original series
diverges.
Example. Does the series converge or diverge?
The series converges, by the Root Test.
Example. Does the series converge
or diverge?
Apply the Root Test:
The series diverges by the Root Test.
Example. Does the series
converge or diverge?
In this case, the Root Test would probably not be a good
choice. Why? Because I'd have on the bottom,
and I don't see an easy way to compute the limit of that expression.
Instead, the factorial suggests using the Ratio Test. The limiting ratio is
Since the limiting ratio is less than 1, the series converges by the
Ratio Test.
Example. Does the series
converge or diverge?
Take the root of the
term:
I need to compute . I'll compute the limit of the
top.
Let . Then
Hence,
Hence,
Therefore,
The limiting ratio is less than 1. Hence, the series converges, by
the Root Test.
Example. Does the series converge or diverge?
Compute the root of the
term:
I need to compute the limit .
Let .
Then
Hence, by L'H\^opital's Rule,
It follows that . Hence, the series converges, by the Root
Test.
Copyright 2019 by Bruce Ikenaga