infinite series and comparison tests tests for infinite series_tcm6-61153.pdfthe basic comparison...

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Infinite Series and Comparison Tests Of all the tests you have seen do far and will see later, these are the trickiest to use because you have to have some idea of what it is you are trying to prove. If a series is divergent and you erroneously believe it is convergent, then applying these tests will lead only to extreme frustration. But there are some pointers that you can use in order to apply them successfully. Here are the two tests we will study: This is intuitively clear; let us look at the convergence case where we have a series which satisfies and where we know that converges. Then the situation with the partial sums, which from a monotonic increasing sequence, is this: Where refers to the partial sums of the series and to that of Being this the case, it follows (not immediately obvious but intuitive) that: The Basic Comparison Test Suppose for all m some nonnegative integer. Then:

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Page 1: Infinite Series and Comparison Tests Tests for Infinite Series_tcm6-61153.pdfThe Basic Comparison Test ... In order to apply this test to a given series we must have a feeling as to

Infinite Series and Comparison Tests

Of all the tests you have seen do far and will see later, these are the trickiest to use because

you have to have some idea of what it is you are trying to prove. If a series is divergent and you

erroneously believe it is convergent, then applying these tests will lead only to extreme

frustration. But there are some pointers that you can use in order to apply them successfully.

Here are the two tests we will study:

This is intuitively clear; let us look at the convergence case where we have a series

which satisfies and where we know that converges. Then the situation

with the partial sums, which from a monotonic increasing sequence, is this:

Where refers to the partial sums of the series and to that of Being this the

case, it follows (not immediately obvious but intuitive) that:

The Basic Comparison Test

Suppose for all m some nonnegative integer. Then:

Page 2: Infinite Series and Comparison Tests Tests for Infinite Series_tcm6-61153.pdfThe Basic Comparison Test ... In order to apply this test to a given series we must have a feeling as to

Thus, if , exists and so must . A similar argument can be used

for the divergence case.

In order to apply this test to a given series we must have a feeling as to whether it

converges or diverges. If we feel it converges, then we need to come up with a sequence

whose terms bound from above ( and whose corresponding

sum .

If, on the other hand, we feel that the give series diverges, then we need to come up with a

sequence whose terms bound from below and whose corresponding sum

.

How do we get this feeling? With practice and through the intuition that practice gives us. But

there are some pointers you can follow.

Ask: what does look like for large n? We will use the symbol to mean that for large

n the terms are of the same order of magnitude or behave more or less the same way. Thus,

means that behaves like a p-series with p = 2 and therefore is expected to

converge. We usually seek p-series to make comparisons because we know under what

circumstances they converge or diverge.

A few examples will illustrate this concept.

Example 1 Determine whether converges or diverges.

Solution

We begin by factoring the largest power of n in order to determine what looks like

for large n:

Page 3: Infinite Series and Comparison Tests Tests for Infinite Series_tcm6-61153.pdfThe Basic Comparison Test ... In order to apply this test to a given series we must have a feeling as to

Thus, for large n

and we expect divergence. Therefore, we seek a sequence with and

such that is known to diverge.

What should we put in place of “?”?

We need to make the fraction smaller. We can achieve this by making its

numerator smaller and/or its denominator larger. Not much we can do with the

numerator, but we know this:

Keep this identity in mind because you will use it substantially.

Thus,

or

From which it follows that

Since diverges (why?), also diverges and we are done.

Page 4: Infinite Series and Comparison Tests Tests for Infinite Series_tcm6-61153.pdfThe Basic Comparison Test ... In order to apply this test to a given series we must have a feeling as to

Example 2 Determine whether converges or diverges.

Solution

Once again, factoring the largest power of n in order to determine what looks like

for large n:

Therefore, we expect convergence and we need to come up with a such that

and such that is known to converge. So now we have to make our fraction larger.

How do we do this? We make the numerator larger and/or the denominator smaller.

Once again, not much we can do with the numerator although is

tempting, but it leads nowhere; we still don’t know what does.

How do we make the denominator smaller? We take more away from :

how about

Thus,

Since (why?), so does .

Page 5: Infinite Series and Comparison Tests Tests for Infinite Series_tcm6-61153.pdfThe Basic Comparison Test ... In order to apply this test to a given series we must have a feeling as to

Example 3 Determine whether converges or diverges.

Solution

The operational thing to observe here is that for which can be seen from the

graph below:

Thus we can write

and we expect convergence. Thus, we seek to bound from above and the choice is

clear in light of the inequality :

Since converges, so does .

Example 4 Determine whether converges or diverges.

Solution

Page 6: Infinite Series and Comparison Tests Tests for Infinite Series_tcm6-61153.pdfThe Basic Comparison Test ... In order to apply this test to a given series we must have a feeling as to

We need to get an idea of how behaves in terms of the series which we know

to converge or diverge. Using the inequality we can write

But this does not help because although we know diverges (why?) we still

don’t know what does. However, we still have a feeling that

diverges; there is enough in the numerator to make it grow fast enough relative to its

denominator. So we need to bound its terms from below with terms whose sum

diverges:

Since for , we can write

for . Now we are getting somewhere. We can also write

And finally, the left-most expression reduces to . Thus,

Since (why?) diverges, so does .

Our next comparison test is a little more mechanical in nature and it is called the Limit

Comparison Test:

The Limit Comparison Test

Suppose and be sequences and

Then,

Page 7: Infinite Series and Comparison Tests Tests for Infinite Series_tcm6-61153.pdfThe Basic Comparison Test ... In order to apply this test to a given series we must have a feeling as to

To use this test given a series we have to come up with a series as our comparing

series. Of course we must know the behavior of , but we can always default to the know

p-series, either using when we suspect divergence or when we suspect

convergence. Here are examples of each case:

Example 5 Determine whether converges or diverges.

Solution

Here which we suspect converges because it has the form for large

n. thus, let

and compute

Simplifying,

Page 8: Infinite Series and Comparison Tests Tests for Infinite Series_tcm6-61153.pdfThe Basic Comparison Test ... In order to apply this test to a given series we must have a feeling as to

Since both series do the same thing. Since converges, so

does .

Example 6 Determine whether converges or diverges.

Solution

This time behaves like so we suspect divergence. Thus, we use

as our comparing sequence:

Since and diverges, so does .

Example 7 Determine whether converges or diverges.

Solution

Let us apply the limit comp test blindly; let be our comparing sequence and let

us see what limit we get:

Page 9: Infinite Series and Comparison Tests Tests for Infinite Series_tcm6-61153.pdfThe Basic Comparison Test ... In order to apply this test to a given series we must have a feeling as to

This is case III of our result: If and diverges, so does .

Since and diverges, so does .

Example 8 Determine whether converges or diverges.

Solution

Once again, we blindly try :

You might be tempted to think that this is case II of our theorem but it is not; the limit is

0, but does not converge. Thus, nothing can be concluded; we should have used

a different comparing sequence.

Let us try :

Page 10: Infinite Series and Comparison Tests Tests for Infinite Series_tcm6-61153.pdfThe Basic Comparison Test ... In order to apply this test to a given series we must have a feeling as to

This is case II: If and converges, so does .

Example 9 Determine whether converges or diverges.

Solution

This series involves the all important which we will encounter time and time again.

Here,

We should ask: What is because this is what we should always do first.

Clearly this is a case of but one in which we cannot use l’Hopital’s rule. How do we

proceed? We use the definition of :

The table below shows how the numerator and denominator behave:

n-values 10^n n!

1 10 1

2 100 2

3 1000 6

4 10000 24

5 100000 120

6 1000000 720

Page 11: Infinite Series and Comparison Tests Tests for Infinite Series_tcm6-61153.pdfThe Basic Comparison Test ... In order to apply this test to a given series we must have a feeling as to

You can see how much quicker grows and it is tempting to conclude

that . However, look at the following table:

It looks totally the opposite of the first table! This is why in math we cannot use

arguments based on observations; if we make an assertion, we must prove it and what

we want to prove here is that, in fact,

Once we have done this, we will prove that converges by either comparison

test.

Exercise: show .

Let us use the comparison test by observing a simpler case:

.

.

.

By the time we get to the 5th term you can see that

n-Values 2^n n!

1 2 1

2 4 2

3 8 6

4 16 24

5 32 120

6 64 720

7 128 5040

Page 12: Infinite Series and Comparison Tests Tests for Infinite Series_tcm6-61153.pdfThe Basic Comparison Test ... In order to apply this test to a given series we must have a feeling as to

and you can verify that

Thus, converges since is a convergent geometric series and

and differ only by a finite number of terms.

Now we reproduce the argument for . Once , we will have the following

situation:

Now we can argue in the same way as the previous case and write:

Thus, the comparison test tells us that converges because

is a convergent geometric series. Since the only difference

and is a finite number of terms, also converges.

n – 9 terms

Page 13: Infinite Series and Comparison Tests Tests for Infinite Series_tcm6-61153.pdfThe Basic Comparison Test ... In order to apply this test to a given series we must have a feeling as to