22.03.2023 - 21:37

Estimate the value of sum^{infty}_{n = 1} \frac{1}{n^{3}}) to within 0.009 of its exact value. (Round to three decimal places as needed.)

Question:

Estimate the value of {eq}sum^{infty}_{n = 1} \frac{1}{n^{3}}) {/eq} to within 0.009 of its exact value. (Round to three decimal places as needed.)

Answers (1)
  • Genevieve
    April 13, 2023 в 04:01
    Using the integral test, we can estimate the value of the given series. Let the integral function be f(x) = (1/x^3). Then, we have: ?f(x)dx = ?(1/x^3)dx = (-1/2x^2) + C Using the limit comparison test, we can compare our series with the p-series (1/n^2): lim (n??) (an/bn) = lim (n??) (1/n^3)/(1/n^2) = lim (n??) 1/n = 0 Since the limit is finite and positive, both series converge or diverge together. The p-series (1/n^2) is a known convergent series, and since the given series is smaller, it also converges. Therefore, using the integral test, we have: ?f(x)dx = (-1/2x^2) + C Thus, lim (n??) (-1/2n^2 + C) - (-1/2) = 1/2 This means that the value of the given series is \approx imately 1/2. To estimate the error, we can use the remainder formula for integral tests: Rn = ?(n+1)? f(x)dx Rn = ?(n+1)? (1/x^3)dx Rn = (-1/2(n+1)^2) Using this formula, we can find that the error is less than 0.009 when: |(-1/2(n+1)^2)| < 0.009 Solving for n, we get: n > 31.1 Therefore, we need to add the terms up until the 32nd term to get an estimate within 0.009 of the exact value. Thus, the estimated value of the series is: ?(from n = 1 to 32) (1/n^3) ? 1.202 (Note that the exact value is actually ?^2/6, which is \approx imately 1.644.)
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