9. Determine the number of terms needed to approximate the sum of the convergent serie. an error of less than 0.001. a. n=0 00 b. Σ n=1 (-1)" 1 2"n! √e = -1)"+1 n² C. (-1)"+1 n=1 n4" = In 5|4
9. Determine the number of terms needed to approximate the sum of the convergent serie. an error of less than 0.001. a. n=0 00 b. Σ n=1 (-1)" 1 2"n! √e = -1)"+1 n² C. (-1)"+1 n=1 n4" = In 5|4
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
I need help with number 9 a to c.
![9.
g. Σ=1(−1)"(Vn+1-vn)
h. Σ=o
a.
Determine the number of terms needed to approximate the sum of the convergent series with
an error of less than 0.001.
n=0
a.
C.
n+1
b. Σ(1)"Η
η
n=1
n=1
n=0
(-3)^
(2n+1)!
(-1)" 1
2"n!
n=1
ΣΤ
n=l
3"
(n+1)"
η
4"
=
10. Use the root or ratio tests to determine the convergence or divergence of the series. In the test
is inconclusive, use another test. State which test(s) you use.
ΣΕΙΡ
g. Σ=1
h. Σ=1
e. Σ=1k(3)
f. Σ=1
cos
√e
ηπ
3
n!
(2η)!
(n!)2
k
C.
2ηη!
5.8.11 (3n+2)
i.
j.
n=1
|.
n=1
n=0
ο. Σ=1(−1)n-1ne-n
∞
κ. Σ
(−1)n+¹
n4"
n=1
(−1)"
n=2(Inn)"
n’ +1
η
n!
(-1)"
24n
' (2n + 1)!
(n!)"
(η)
-
8
m. Ln=1
In
sin 4η
4η
( ( 1 + ² ) ¹²
η
2η2
n!
n. Σ=1
ο. Σ=1
5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa098e07d-b15e-4284-9177-08ea985b9919%2F336aa116-9ee0-4aa4-9461-88a69fdfe93c%2F79nlgrb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:9.
g. Σ=1(−1)"(Vn+1-vn)
h. Σ=o
a.
Determine the number of terms needed to approximate the sum of the convergent series with
an error of less than 0.001.
n=0
a.
C.
n+1
b. Σ(1)"Η
η
n=1
n=1
n=0
(-3)^
(2n+1)!
(-1)" 1
2"n!
n=1
ΣΤ
n=l
3"
(n+1)"
η
4"
=
10. Use the root or ratio tests to determine the convergence or divergence of the series. In the test
is inconclusive, use another test. State which test(s) you use.
ΣΕΙΡ
g. Σ=1
h. Σ=1
e. Σ=1k(3)
f. Σ=1
cos
√e
ηπ
3
n!
(2η)!
(n!)2
k
C.
2ηη!
5.8.11 (3n+2)
i.
j.
n=1
|.
n=1
n=0
ο. Σ=1(−1)n-1ne-n
∞
κ. Σ
(−1)n+¹
n4"
n=1
(−1)"
n=2(Inn)"
n’ +1
η
n!
(-1)"
24n
' (2n + 1)!
(n!)"
(η)
-
8
m. Ln=1
In
sin 4η
4η
( ( 1 + ² ) ¹²
η
2η2
n!
n. Σ=1
ο. Σ=1
5
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