Use the comparison or limit comparison test to decide if the

kerrum75

kerrum75

Answered question

2021-12-22

Use the comparison or limit comparison test to decide if the following series converge.
n=14sinn2n+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

SlabydouluS62

SlabydouluS62

Skilled2021-12-23Added 52 answers

n=14sinn2n+1
Since 4sinn2n+1<32n (since the maximum value of sinn is 1)
So n=14sinn2n+1<n=132n
Since n=132n is convergent so n=14sinn2n+1 is convergent
So the approximate value
n=132n=3(12+122+123)+
=3{(1+12+122+123)1}
=3{(112)11}
=3{(12)11}
=3(21)
=3
chumants6g

chumants6g

Beginner2021-12-24Added 33 answers

Use the root test to determine the convergence of the series:
n4sin(n)2n+1
Recall the statement of the root test.
The convergence of a series kak can be determined by examining the quantity ρ=limk|ak|1k using the following comparisons:
If ρ<1, the series converges absolutely.
If ρ>1 or ρ=, the series diverges.
If ρ=1, the series may converge or diverge and the test is inconclusive.
Check the convergence of the series by taking the limit of |4sin(n)2n+1|1n.
Take the limit of the n-th root of the absolute value of the n-th term.
limn|4sin(n)2n+1|1n=12
Compare the result to 1.
Since the limit is less than 1, the series n4sin(n)2n+1 converges absolutely:
Answer: The series is absolutely convergent by the root test.
user_27qwe

user_27qwe

Skilled2021-12-30Added 375 answers

The first answer is clearer

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