Use the comparison or limit comparison test to decide if the following series converge.
Series from n=1 to infinity of (4-sin n) / ((n^2)+1) and the series from n=1 to infinity of (4-sin n) / ((2^n) +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" the sum from n=6 to infinity of an, by comparing it with an appropriate improper integral or geometric series.
Can anyone just give me an idea on how to go about solving this problem
someone help!!!!
i know they are both convergent but idk how to do the second part of the problem
How'd you find out that they're both convergent?
through the comparison test
hey lauren did you find anything? im still lost. please help me if you find the answer.
To decide if the first series converges, we can use the comparison test.
Step 1: Getting the intuition
First, let's consider the behavior of the term (4 - sin(n)) / ((n^2) + 1) as n approaches infinity. The numerator, 4 - sin(n), is always between 3 and 5. Therefore, we can say that the term is bounded by [3 / ((n^2) + 1)] and [5 / ((n^2) + 1)].
Step 2: Comparing with a known series
We can compare this series with the p-series ∑ (1 / (n^2)). The p-series converges when p > 1, which is the case here (p = 2). Therefore, if our series is less than or equal to a converging series, it must also converge.
Step 3: Applying the comparison test
Using the comparison test, we can write:
0 ≤ (4 - sin(n)) / ((n^2) + 1) ≤ (5 / ((n^2) + 1))
The series ∑ (5 / ((n^2) + 1)) can be rewritten as 5 * (∑ (1 / ((n^2) + 1))). By comparing it with the p-series ∑ (1 / (n^2)), we find that it converges.
Therefore, by the comparison test, the series ∑ (4 - sin(n)) / ((n^2) + 1) also converges.
To find an approximation of the sum and an error estimate, we'll calculate s5 (the sum of the first 5 terms) and estimate the tail sum from n = 6 to infinity.
s5 = (4 - sin(1)) / (1^2 + 1) + (4 - sin(2)) / (2^2 + 1) + (4 - sin(3)) / (3^2 + 1) + (4 - sin(4)) / (4^2 + 1) + (4 - sin(5)) / (5^2 + 1)
Now, for the tail sum estimation, we can compare it with an improper integral. Let's choose the integral:
∫[(4 - sin(x)) / (x^2 + 1)]dx from 5 to infinity.
Calculating the integral may not be straightforward, so we can use software or numerical methods to approximate it. Let's assume the result is A.
We can estimate the tail sum as follows:
Tail sum ≈ A = ∫[(4 - sin(x)) / (x^2 + 1)]dx from 5 to infinity
By subtracting s5 from A, we can obtain an approximation for the tail sum. Keep in mind that this is an approximation and the actual tail sum may be slightly different.
Repeat the above steps for the second series:
Series: ∑ (4 - sin(n)) / ((2^n) + 1)
We can use the same approach, comparing it with the converging geometric series ∑ (1 / (2^n)).
Following the steps, we can conclude that the second series also converges.
I hope this explanation helps you understand the process of using the comparison test and estimating the sum and error for a given series.