Use the comparison or limit comparison test to decide if the

Irvin Dukes

Irvin Dukes

Answered question

2021-12-26

Use the comparison or limit comparison test to decide if the following series converge.
n=14sinnn2+1
For each series which converges, give an approximation of its sum, together with an error estimate, as follows. First calculate the sum s5 of the first 5 terms, then estimate the "tail" n=6an by comparing it with an appropriate improper integral or geometric series.

Answer & Explanation

Timothy Wolff

Timothy Wolff

Beginner2021-12-27Added 26 answers

Given series is n=14sinnn2+1
Since 4sinnn2+1<3n2 (since the maximum value of sinn is 1)
And n=13n2 is convergent
Since n=14sinnn2+1<n=13n2 and n=13n2 is convergent
By comparison test n=14sinnn2+1 is convergent
psor32

psor32

Beginner2021-12-28Added 33 answers

Use the limit test to determine the convergence of the series:
n4sin(n)n2+1
Recall the statement of the limit test.
The convergence of a series nan can be determined by examining the quantity ρ=limnan using the following comparisons:
If |ρ|>0 or if ρ is undefined, the series diverges.
If ρ=0, the limit test is inconclusive.
Check the divergence of the series by computing the limit of the summand.
Take the limit of the summand as n approaches :
limn4sin(n)n2+1=0
INTERMEDIATE STEPS:
Find the following limit:
limn4sin(n)n2+1
A bounded function times one that approaches 0 also approaches 0.
Since 34sin(n)5 for all n, and since limn1n2+1=0, we have that limn4sin(n)n2+1=0:
Answer:0
If the limit of the summand as the iterator n goes to is nonzero, then the series must diverge.
Since the limit is equal to 0, the limit test is inconclusive:
Answer: The limit test is inconclusive.
user_27qwe

user_27qwe

Skilled2021-12-30Added 375 answers

The second answer is more understandable

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