Welcome :: Homework Help and Answers :: Mathskey.com

Recent Visits

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

13,435 questions

17,804 answers

1,438 comments

776,824 users

help? comparision[integral]??

0 votes

asked Jun 1, 2015 in CALCULUS by anonymous

3 Answers

0 votes

(1)

Step 1:

The series is image.

The 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 1 and the dominant part of the denominator is image.

Observe that image.

Now compare the given series with the series image.

Step 2 :

The obtained series is image.

Definition of p - series :

The p - series is convergent if and divergent if .

Here p = 1.

Hence the series image is divergent.

Here image and image is divergent.

By comparison test, image also diverges.

Solution :

image is divergent.

answered Jun 1, 2015 by Lucy Mentor
0 votes

(2)

Step 1 :

The series is .

The 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 n.

Observe that .

Now compare the given series with the series image.

Step 2 :

The obtained series is image.

Definition of p - series :

The p - series is convergent if and divergent if .

Here p = 1.

Hence the series image is divergent.

Here and image is divergent.

By comparison test, also diverges.

Solution :

is divergent.

answered Jun 1, 2015 by Lucy Mentor
0 votes

(3)

Step 1:

The series is image.

Integral test :

Suppose image is continuous , positive and decreasing on the interval image then

(i) image is convergent, then the image is also convergent.

(ii) image is divergent, then the image is also divergent.

The function image is continuous , positive and decreasing on the interval image.

image

image

image

image diverges, hence by integral test image also diverges.

The series is divergent.

Solution :

The series is divergent.

answered Jun 1, 2015 by Lucy Mentor
edited Jun 1, 2015 by Lucy

Related questions

asked May 16, 2015 in CALCULUS by anonymous
asked Oct 23, 2014 in CALCULUS by anonymous
asked Jul 10, 2014 in CALCULUS by anonymous
...