Limited time offer 20% OFF StudySoup Subscription details

Utah State University - MATH 1220 - Class Notes - Week 5

Created by: Colton Hill Elite Notetaker

> > > > Utah State University - MATH 1220 - Class Notes - Week 5

Utah State University - MATH 1220 - Class Notes - Week 5

School: Utah State University
Department: Applied Mathematics
Course: Calculus 2
Professor: Greg Wheeler
Term: Spring 2019
Tags: Math and Calculus
Name: Math HW 8 & 9 Solutions (With work shown)
Description: These notes cover the series and sequences. 8 is primarily finding the limit of the problem, 9 is finding the convergence of the summation of series.
Uploaded: 02/06/2019
This preview shows pages 1 - 3 of a 7 page document. to view the rest of the content
background image Colton Hill
February 7 2019
Math 1220-502
Homework Problems 9 1. Determine the first six partial sums of each series. Then determine the sum of the series or show that the series diverges. (a) n=1 2 3 n−1 n=1 2 1 3 n−1 = n=1 2  1 3 n−1 , r < 1 n=1 2 1 3 n−1 = 2 1 1 3 = 3 a 1 = 2, a 2 = 2
3
, a 3 = 2 3 2 , a 4 = 2 3 3 , a 5 = 2 3 4 , a 6 = 2 3 5 Solution: s 1 = 2 s 2 = 8
3
s 3 = 26 9 s 4 = 80
27
s 5 = 242 81 s 6 = 728
243
The series converges to 3. (b) n=1 9 100 n n=1 9 100 n = n=1 9 100 1 100 n−1 , r < 1 n=1 9 100 1 100 n−1 = 9 100 1 1 100 = 1 11 a 1 = 0.09, a 2 = 0.0009, a 3 = 0.000009, a 4 = 0.00000009, a 5 = 0.0000000009, a 6 = 0.000000000009 Solution: s 1 = 0.09 s 2 = 0.0909 s 3 = 0.090909 s 4 = 0.09090909 s 5 = 0.0909090909 s 6 = 0.090909090909 The series converges to 1 11 . (c) n=1 (− 1 ) n 5 4 n−1 n=1 (− 1 ) n 5 4 n−1 = n=1 (− 5 ) 1 4 n−1 , | r | < 1 n=1 (− 5 ) 1 4 n−1 = 5 1 + 1 4 = − 4 a 1 = − 5, a 2 = 5
4
, a 3 = − 5 16 , a 4 = 5 64 , a 5 = − 5 256 , a 6 = 5 1024 Solution: s 1 = − 5 s 2 = − 15 4 s 3 = − 65
16
s 4 = − 255 64 s 5 = − 1025 256 s 6 = − 4095
1024
The series converges to 4. (d) n=1 (− 3 ) n−1 2 n n=1 (− 3 ) n−1 2 n = n=1 1
2
3 2 n−1 . | r | > 1 it diverges a 1 = 1
2
, a 2 = 3 4 , a 3 = 9
8
, a 4 = 27 16 , a 5 = 81
32
, a 1 = 243 64 Solution: s 1 = 1
2
s 2 = − 1
4
s 3 = 7
8
s 4 = − 13
16
s 5 = 55
32
s 6 = − 133 64 The series diverges. 1
background image (e) n=1  2 3 n + 1 3 n−1 n=1  2 3 n + 1 3 n−1 = n=1  2 3 1 3 n−1 + 1 3 n−1 = n=1 5
3
 1 3 n−1 , r < 1 n=1  2 3 n + 1 3 n−1 = 5 3 1 1 3 = 5
2
a 1 =  2 3 + 1 , a 2 =  2 9 + 1
3
, a 3 =  2 27 + 1
9
, a 4 =  2 81 + 1 27 , a 5 = 2 243 + 1 81 , a 6 = 2 729 + 1 243 Solution: s 1 = 5
3
, s 2 = 20 9 , s 3 = 65
27
, s 4 = 200 81 , s 5 = 605
243
, s 6 = 1820 729 The series converges to 5
2
. (f) n=1 1 1 2 n−1 Using the Divergence Test: lim n→ ( a n ) = lim n→ 1 1 2 n−1 = 1 it divergence a 1 = 0, a 2 = 1
2
, a 3 = 3
4
, a 4 = 7
8
, a 5 = 15
16
, a 6 = 31
32
Solution: s 1 = 0 s 2 = 1
2
s 3 = 5
4
s 4 = 17 8 s 5 = 49
16
s 6 = 129 32 The series diverges. (g) n=1 1 n + 3 1 n + 4 a 1 =  1 4 1
5
, a 2 =  1 5 1
6
, a 3 =  1 6 1
7
, a 4 =  1 7 1
8
, a 5 =  1 8 1
9
, a 6 =  1 9 1 10 This is a telescoping with all terms canceling out save ( 1
4
1 ) Solution: s 1 = 1 20 s 2 = 1 12 s 3 = 3 28 s 4 = 1
8
s 5 = 5 36 s 6 = 3 20 The series converges to 1
4
. (h) n=1 3 n 2 + n Hint: The sum will be easier to find if you rewrite the sequence using partial fractions. n=1 3 n 2 + n = n=1 A n + B n + 1 3 = A ( n + 1 ) + Bn 0 = A + b, 3 = A, B = − 3 n=1 3 n 2 + n = n=1 3 n + 3 n + 1 a 1 = 3 + 3 2 , a 2 = 3
2
1, a 3 = 1 + 3 4 , a 4 = 3
4
+ 3 5 , a 5 = 3
5
+ 1 2 , a 6 = 1
2
+ 3 7 This sequence is telescoping n=1 3 n 2 + n = 3 3 Solution: s 1 = 3
2
s 2 = 2 s 3 = 9
4
s 4 = 12 5 s 5 = 5
2
s 6 = 18 7 The series converges to 3. (i) n=1 n + 1 2n 1 a 1 = 2
1
, a 2 = 3
3
, a 3 = 4
5
, a 4 = 5
7
, a 5 = 6
9
, a 6 = 7 11 lim n→ n + 1 2n 1 = 1
2
It Divergences Solution: s 1 = 2 s 2 = 3 s 3 = 19 5 s 4 = 158 35 s 5 = 544
105
s 6 = 6719
1155
The series diverges. 2
background image 2. What does the divergence test tell us about the convergence/divergence for each of the following series? (a) n=1 1 2n 1 lim n→ 1 2n 1 = 0 Solution: The divergence test is inconclusive. (b) n=1 ( 2n + 3 ) lim n→ ( 2n + 3 ) = Solution: The series diverges. (c) n=1 3n 2 + 2n 1 n + 2 lim n→ 3n 2 + 2n 1 n + 2 = Solution: The series diverges. (d) n=1 3n 2 + 2n 1 n 2 + 2 lim n→ 3n 2 + 2n 1 n 2 + 2 = 3 Solution: The series diverges. (e) n=1 3n 2 + 2n 1 10000n 2 + 2 lim n→ 3n 2 + 2n 1 10000n 2 + 2 = 0.0003 Solution: The series diverges. (f) n=1 3n 2 + 2n 1 n 3 + 2 lim n→ 3n 2 + 2n 1 n 3 + 2 = 0 Solution: The divergence test is inconclusive. (g) n=1 | sin n | lim n→ | sin n | = ( 0 to 1 ) Solution: The series diverges. (h) n=1 sin  1 n lim n→ sin  1 n = sin ( 0 ) = 0 Solution: The divergence test is inconclusive. 3

This is the end of the preview. Please to view the rest of the content
Join more than 18,000+ college students at Utah State University who use StudySoup to get ahead
7 Pages 65 Views 52 Unlocks
  • Better Grades Guarantee
  • 24/7 Homework help
  • Notes, Study Guides, Flashcards + More!
Join more than 18,000+ college students at Utah State University who use StudySoup to get ahead
School: Utah State University
Department: Applied Mathematics
Course: Calculus 2
Professor: Greg Wheeler
Term: Spring 2019
Tags: Math and Calculus
Name: Math HW 8 & 9 Solutions (With work shown)
Description: These notes cover the series and sequences. 8 is primarily finding the limit of the problem, 9 is finding the convergence of the summation of series.
Uploaded: 02/06/2019
7 Pages 65 Views 52 Unlocks
  • Better Grades Guarantee
  • 24/7 Homework help
  • Notes, Study Guides, Flashcards + More!
Join StudySoup for FREE
Get Full Access to Utah State University - Class Notes - Week 5
Join with Email
Already have an account? Login here
×
Log in to StudySoup
Get Full Access to Utah State University - Class Notes - Week 5

Forgot password? Reset password here

Reset your password

I don't want to reset my password

Need help? Contact support

Need an Account? Is not associated with an account
Sign up
We're here to help

Having trouble accessing your account? Let us help you, contact support at +1(510) 944-1054 or support@studysoup.com

Got it, thanks!
Password Reset Request Sent An email has been sent to the email address associated to your account. Follow the link in the email to reset your password. If you're having trouble finding our email please check your spam folder
Got it, thanks!
Already have an Account? Is already in use
Log in
Incorrect Password The password used to log in with this account is incorrect
Try Again

Forgot password? Reset it here