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

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

##### 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.
This preview shows pages 1 - 3 of a 7 page document. to view the rest of the content 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 (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 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

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##### 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.
7 Pages 63 Views 50 Unlocks
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