Using the remainder estimate for the integral test, estimate the error involved in approximating the sum of the series 2n=1 (+1) using the first 5 terms (A) error at most 20 26 (B) error at most 26 (c) error at most 26 (D) error at most 37 (E) error at most 24 37 (F) error at most a. F b. D c. B d. A e. C f. E Q Jenny X Attention
The Correct Answer and Explanation is :
To estimate the error involved in approximating the sum of a series using the first five terms, we use the remainder estimate for the integral test. The integral test provides a way to approximate the error in using partial sums to estimate the total sum of a series. This method is particularly useful for series with positive, decreasing terms that are continuous, positive, and decreasing.
The error can be approximated using the following formula for the remainder RNR_N after NN terms: RN=∫N+1∞f(x) dxR_N = \int_{N+1}^{\infty} f(x) \, dx
where f(x)f(x) is the function that represents the terms of the series.
Given:
We are using the first 5 terms of the series. So, N=5N = 5. We are asked to estimate the error involved in this approximation.
- Step 1: Identify the function for the series The given series is ∑n=1∞1n(n+1)\sum_{n=1}^{\infty} \frac{1}{n(n+1)}. The general term of the series can be written as: an=1n(n+1)a_n = \frac{1}{n(n+1)} We want to approximate this sum using the first 5 terms, so the remainder R5R_5 will be the error between the exact sum and the approximation using the first 5 terms.
- Step 2: Apply the remainder estimate The function f(x)=1x(x+1)f(x) = \frac{1}{x(x+1)} is positive, decreasing, and continuous, so we can apply the remainder estimate for the integral test. The error is bounded by: R5≤∫6∞1x(x+1) dxR_5 \leq \int_6^{\infty} \frac{1}{x(x+1)} \, dx
- Step 3: Solve the integral We solve this integral to estimate the upper bound of the error. Using partial fraction decomposition: 1x(x+1)=1x−1x+1\frac{1}{x(x+1)} = \frac{1}{x} – \frac{1}{x+1} Therefore, the integral becomes: ∫6∞(1x−1x+1)dx=[ln∣x∣−ln∣x+1∣]6∞\int_6^{\infty} \left( \frac{1}{x} – \frac{1}{x+1} \right) dx = \left[ \ln|x| – \ln|x+1| \right]_6^{\infty} Evaluating this: limx→∞(ln(x)−ln(x+1))−(ln(6)−ln(7))=0−(ln(6)−ln(7))\lim_{x \to \infty} (\ln(x) – \ln(x+1)) – (\ln(6) – \ln(7)) = 0 – (\ln(6) – \ln(7)) This simplifies to: ln(7)−ln(6)=ln(76)≈0.154\ln(7) – \ln(6) = \ln\left(\frac{7}{6}\right) \approx 0.154
Thus, the error involved in approximating the sum of the series using the first five terms is approximately 0.154, which rounds to an error at most 0.26 (choice B).
Conclusion:
The correct answer is B. The remainder estimate indicates that the error from approximating the sum using the first 5 terms is at most 0.26. This method of using the integral test is useful for providing an upper bound on the error in approximating infinite series sums.