consider the power series "(x-2)"" √n+2 n=o 1. Use Limit Comparison Test to show that the series (-3)^(x-2)^ √n+2 (-3)^(-2)^ √n+2 (-3)^(-1)^ n+2 x=¾: 2 11 11 n³0 10 no √n+2 +7

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section: Chapter Questions
Problem 22RE
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(-3)^(x-2)^
Consider the power series.
√n+2
n=o
1. Use Limit Comparison Test to show that the series diverges when x=-=-
X=
x = 3:
(-3)^(x-2)^
√n+2
(-3)^(-2)^
=
√n+2
(-3)^(-1)^
√n +2
at x =
n=0
⇒diverges
lim
1 > 0
- 13/12
Now, the series is divergent at x = -
Since an diverges, therefore by Limit Comparison Test, bn also diverges.
348
=
Now, bn
=
n
Hence, an <bn
Using Limit Comparison Test
√√n+2
lim an
lim
0+2
=
n+f
bo
عالم
=
10
no √n+2
√nt2
n+ 2
n=o
lim
n+ (1)
√n+2
nt 2
Transcribed Image Text:(-3)^(x-2)^ Consider the power series. √n+2 n=o 1. Use Limit Comparison Test to show that the series diverges when x=-=- X= x = 3: (-3)^(x-2)^ √n+2 (-3)^(-2)^ = √n+2 (-3)^(-1)^ √n +2 at x = n=0 ⇒diverges lim 1 > 0 - 13/12 Now, the series is divergent at x = - Since an diverges, therefore by Limit Comparison Test, bn also diverges. 348 = Now, bn = n Hence, an <bn Using Limit Comparison Test √√n+2 lim an lim 0+2 = n+f bo عالم = 10 no √n+2 √nt2 n+ 2 n=o lim n+ (1) √n+2 nt 2
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