Hid Apply Series Comparison Test: diverges Can someone explain how an >= bn for all n? I tried substituting values for n but it seems that 5" un=1 3" – |sin(n)| - an < bn Comparison Test: Hide Def Let an, Lb, be two positive sequences such that for all n, An < bn If E b, converges, so does an IfΣ α, diverges, so does Σb

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.2: Arithmetic Sequences
Problem 67E
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Apply Series Comparison Test: diverges
5"
Can someone explain how an >= bn
for all n? I tried substituting values for
un=1
3" – |sin(n)|
n but it seems that
an < bn
Comparison Test:
Hide Definition
Let an, L b, be two positive sequences such that for all n,
an< bn
If E b, converges, so does a,
If E a, diverges, so does b,
5"
5"
-1< sin(n)< 1, 0s |sin(n)|< 1
Σ
3" – |sin(n)|
3" – 1
Show Steps
5"
Check convergence of Ln=1
3"
diverges
1
Transcribed Image Text:Hide Steps Apply Series Comparison Test: diverges 5" Can someone explain how an >= bn for all n? I tried substituting values for un=1 3" – |sin(n)| n but it seems that an < bn Comparison Test: Hide Definition Let an, L b, be two positive sequences such that for all n, an< bn If E b, converges, so does a, If E a, diverges, so does b, 5" 5" -1< sin(n)< 1, 0s |sin(n)|< 1 Σ 3" – |sin(n)| 3" – 1 Show Steps 5" Check convergence of Ln=1 3" diverges 1
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