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
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
Section10.3: Geometric Sequences
Problem 40E
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please show step by step solution on how to get the circled part with complete explanation. do this typewritten and i will upvote. skip if you already did this
![(-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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F23e1e3d5-f4a9-424f-9ca9-933f11816050%2Ff97f535b-11cd-431f-8d4a-6f159dc2c73a%2Fktzrs4s_processed.png&w=3840&q=75)
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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