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.

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.

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.

Hence, we get

and

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

and

and is summable.

is not summable since

and is not 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.

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.

- (a)
- (b)
- (c)
- (d)
- (e)
- (f)
- (g)
- (h)

- 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.

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.

so

Since is not summable, is also not summable.

- a)
- .
- b)
- .

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.

If
for all
, then
for all and hence

, and is summable.
These examples show that when
the
ratio test gives no useful information.

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 .

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

- a)
- b)
- c)
- d)
- e)
- f)
- g) [For this series, there is one for which you don't need to answer the question.]