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# 11.2 Convergence Tests

In this section we prove a number of theorems about convergence of series of real numbers. Later we will show how to use these results to study convergence of complex sequences.

11.11   Theorem (Comparison test for series.) Let be two sequences of non-negative numbers. Suppose that there is a number such that

Then

and

Proof: Note that the two statements in the conclusion are equivalent, so it is sufficient to prove the first.

Suppose that is summable. Then converges, so is bounded -- say for all . Then for all ,

Since for we have

we see that is bounded by . Also is increasing, since . Hence is bounded and increasing, and hence converges; i.e., is summable.

11.12   Examples. Since

and diverges, it follows that also diverges. Since converges for , also converges for .

In order to use the comparison test, we need to have some standard series to compare other series with. The next theorem will provide a large family of standard series.

11.13   Theorem. Let . Then is summable if , and is not summable if .

Proof: Let for . Then for all and all ,

Hence,

If , then , so and is positive. Hence ; i.e., the sequence is bounded. It is also increasing, so it converges.

If , then , so by using the comparison test with the harmonic series, is not summable.

11.14   Remark. For , the proof of the previous theorem shows that

Hence, we get

and

The exact values of the series (found by Euler) are

and

11.15   Examples. is summable, since

and is summable.

is not summable since

and is not summable.

11.16   Example. Let , and let . Then is summable.

Proof: By the reverse triangle inequality, we have for all

so

Since is a summable sequence for all , it follows from the comparison test that is summable.

11.17   Example. Let for all . Then

If , then , so

Hence,

Hence, (by induction)

where ; i.e., for all . Since the geometric series converges, it follows from the comparison test that converges also.

11.18   Exercise. Determine whether or not the sequences below are summable:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)

11.19   Exercise. A Give examples of the following, or explain why no such examples exist.
a)
Two real sequences and such that and are not summable, but is summable.
b)
Two real sequences and such that and are summable, but is not summable.
c)
Two real sequences and such that for all , and is summable but is not summable.

11.20   Theorem (Limit comparison test.) Let be sequences of positive numbers. Suppose that converges to a non-zero limit . Then is summable if and only if is summable.

Proof: We know that . Let be a precision function for . Then

i.e.,

If is summable, then is summable, and since for all , it follows from the comparison test that is summable. If is not summable, then since for all it follows that is not summable, and hence is not summable.

11.21   Example. Is summable? Let . Note that for all . For large , is  like" , so I'll compare this series with . Let for all . Then

so

Since is not summable, is also not summable.

11.22   Exercise. Determine whether or not the sequences below are summable.
a)
.
b)
.

11.23   Theorem (Ratio test.) Let be a sequence of positive numbers. Suppose the converges, and . Then, if , is summable. If , is not summable. (If , the theorem makes no assertion.)

Proof: Suppose .

Case 1: . Let be a precision function for . Then for all ,

Write and , so . Then

so

and (by an induction argument which I omit)

or

or

Since is a summable geometric series, it follows from the comparison test that is also summable.

Case 2: . As before, let be a precision function for . Then for all ,

Hence is not a null sequence. So is not summable.

11.24   Warning. The ratio test does not say that if for all , then is summable. If for , then for all but is not summable. (In this case, , and the ratio test does not apply.)

If for all , then for all and hence
, and is summable. These examples show that when the ratio test gives no useful information.

11.25   Remark. If, in applying the ratio test, you find that for all large , you can conclude that diverges (even if does not exist), since this condition shows that is not a null sequence.

11.26   Example. Let be a positive number and let . We apply the ratio test to the series .

Note that

Hence

From this we see that . The ratio test says that if (i.e., if ), then is summable, and if , then the sequence is not summable.

Can we figure out what happens in the case ? For , our formula above gives us

i.e., for . Thus,

and (by induction),

Since is not summable, it follows that is not summable for .

11.27   Example. Let for all . I'll apply ratio test to . For all ,

Hence and the ratio test does not apply. But since for all , I conclude that is an increasing sequence and hence diverges.

11.28   Exercise. For each of the series below, A determine for which the series converges.
a)
b)
c)
d)
e)
f)
g) [For this series, there is one for which you don't need to answer the question.]

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