a. for the series ∑(n = 1 to ∞) 6 / n³, use the inequalities shown below with n = 10 to find an interval…

a. for the series ∑(n = 1 to ∞) 6 / n³, use the inequalities shown below with n = 10 to find an interval containing the sum s. sn+∫(n + 1 to ∞) f(x) dx≤s≤sn+∫(n to ∞) f(x) dx b. use the mid - point of the interval found in part (a) to approximate the sum of the series. a. ≤∑(n = 1 to ∞) 6 / n³≤ (round to four decimal places as needed.)

a. for the series ∑(n = 1 to ∞) 6 / n³, use the inequalities shown below with n = 10 to find an interval containing the sum s. sn+∫(n + 1 to ∞) f(x) dx≤s≤sn+∫(n to ∞) f(x) dx b. use the mid - point of the interval found in part (a) to approximate the sum of the series. a. ≤∑(n = 1 to ∞) 6 / n³≤ (round to four decimal places as needed.)

Answer

Explanation:

Step1: Define the function and find (s_N)

The function (f(x)=\frac{6}{x^{3}}), and (s_N=\sum_{n = 1}^{N}\frac{6}{n^{3}}). For (N = 10), (s_{10}=6\sum_{n=1}^{10}\frac{1}{n^{3}}=6\left(1+\frac{1}{8}+\frac{1}{27}+\frac{1}{64}+\frac{1}{125}+\frac{1}{216}+\frac{1}{343}+\frac{1}{512}+\frac{1}{729}+\frac{1}{1000}\right)\approx6\times1.197532 = 7.1852).

Step2: Calculate (\int_{N + 1}^{\infty}f(x)dx)

(\int_{11}^{\infty}\frac{6}{x^{3}}dx=\lim_{b\rightarrow\infty}\int_{11}^{b}\frac{6}{x^{3}}dx=\lim_{b\rightarrow\infty}\left[- \frac{3}{x^{2}}\right]{11}^{b}=\lim{b\rightarrow\infty}\left(-\frac{3}{b^{2}}+\frac{3}{121}\right)=\frac{3}{121}\approx0.0248).

Step3: Calculate (\int_{N}^{\infty}f(x)dx)

(\int_{10}^{\infty}\frac{6}{x^{3}}dx=\lim_{b\rightarrow\infty}\int_{10}^{b}\frac{6}{x^{3}}dx=\lim_{b\rightarrow\infty}\left[- \frac{3}{x^{2}}\right]{10}^{b}=\lim{b\rightarrow\infty}\left(-\frac{3}{b^{2}}+\frac{3}{100}\right)=\frac{3}{100}=0.03).

Step4: Find the interval for (S)

Using (s_N+\int_{N + 1}^{\infty}f(x)dx\leq S\leq s_N+\int_{N}^{\infty}f(x)dx), we have (7.1852+0.0248\leq S\leq7.1852 + 0.03), so (7.2100\leq\sum_{n = 1}^{\infty}\frac{6}{n^{3}}\leq7.2152).

Step5: Approximate the sum using the mid - point

The mid - point of the interval ([7.2100,7.2152]) is (\frac{7.2100 + 7.2152}{2}=7.2126).

Answer:

a. (7.2100\leq\sum_{n = 1}^{\infty}\frac{6}{n^{3}}\leq7.2152) b. (7.2126)