Calculus-Sec-9-3-Solutio

9.3 Alternating Series and Absolute Convergence

by SGLee - HSKim - SWSun

1-7. Test for convergence of the following alternating series:

1.

Since  is divergent, the series is divergent by the Integral Test.

2.

Note that  iff

By Alternating Series Test, the series is divergent.

3.

It satisfies  because  and .

By the Alternating Series Test, the series is convergent.

5.

By the Integral Test,

is divergent.

By Alternating Series Test, the series is divergent.

6.

By the Ratio Test, this series is absolutely convergent.

Hence,  is convergent.

7.

By Alternating Series Test, the series is divergent.

8. For what values of , the series  is convergent.

By Exercise 5, when , this series diverges.

Let  and .

For  for all x.

And for

Thus, by the Alternating Series Test, for , this series is convergent. If , this series is divergent.

9-14. Test whether the series is absolutely convergent, conditionally convergent, or divergent.

9.

Hence, the series is divergent.

10.

Divergent by the Ratio test.

11.

.

Since  for all , we have .

Thus

Let

is absolutely convergent.

By the Comparison Test,

is absolutely convergent, and hence convergent.

12.

Using the root test,

is absolutely convergent, and hence convergent.

13.

.

We do not know if this series is absolutely convergent, when using the ratio test. Let us try another      test.

.

Here,  for all  and

By the Alternating Series Test,

is conditionally convergent.

14.

Then

This means we cannot conclude convergence of this series using the Ratio Test.

Consider

Here,  for all  and

By the Alternating Series Test,

is conditionally convergent.

Massachusetts Institute of Technology

(MIT), Boston/Cambridge, MA, USA