Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,459 questions

17,854 answers

1,446 comments

805,839 users

???????????

0 votes

asked Jun 3, 2015 in CALCULUS by anonymous

2 Answers

0 votes

(c)

Step 1:

The series is image.

The Ratio Test:

Let  image be a series with nonzero terms.

1)image converges absolutely if image.

2)image diverges if image and image.

3) The Ratio Test is inconclusive if image.

Step 2:

Consider image.

Apply Ratio Test.

image

This series is convergent because image is less than image.

Solution:

The series image is convergent.

answered Jun 3, 2015 by Lucy Mentor
0 votes

(d)

Step 1 :

The series is image.

Comparison Test :

Suppose that and are series with positive terms.

(i) If is convergent and for all n , then is also convergent.

(ii) If is divergent and for all n, then is also divergent.

The dominant part of the numerator is image and the dominant part of the denominator is image.

Observe that image.

Now the obtained series is image.

Step 2 :

The series is image.

image

Test of divergence : If image doesnot exist or image, then the series image is divergent.

image

By the test of divergence, the series image is divergent.

Here image and image is divergent.

By comparison test, image is also divergent.

Solution :

The series image is divergent.

answered Jun 3, 2015 by Lucy Mentor
edited Jun 3, 2015 by Lucy

Related questions

asked Jul 10, 2015 in CALCULUS by anonymous
asked Jun 3, 2015 in CALCULUS by anonymous
asked Jun 3, 2015 in CALCULUS by anonymous
asked Jun 3, 2015 in CALCULUS by anonymous
asked Jun 3, 2015 in CALCULUS by anonymous
asked Jun 3, 2015 in CALCULUS by anonymous
asked Jun 1, 2015 in CALCULUS by anonymous
...