Chapter 8.3, Problem 38E

To determine

To find: The sum of the given series using the first $10$ terms and to then estimate the error.

Answer to Problem 38E

The estimated of the given series is $0.83253$ and error is less than $0.005$ .

Explanation of Solution

Given information: Given series is ${\displaystyle \sum _{n=1}^{\infty }\frac{{\sin }^{2}n}{n^3}}$ .

Calculation:

The sum of first $10$ terms of the series is:

  $s_{10}=a_1+a_2+\mathrm{...}+a_{10}={\sin }^{2}1+\frac{{\sin }^{2}2}{2^3}+\mathrm{...}+\frac{{\sin }^{2}10}{{10}^3}\approx 0.83253$

According to the Remainder Estimate,

  $\begin{array}{l}{R}_n=s-s_n\le {\displaystyle \underset{n}{\overset{\infty }{\int }}f(x)dx}\\ \\ \Rightarrow R_{10}=s-s_{10}\le {\displaystyle \underset{10}{\overset{\infty }{\int }}\frac{{\sin }^{2}x}{x^3}dx}\\ \\ \le {\displaystyle \underset{10}{\overset{\infty }{\int }}\frac{1}{x^3}dx\left(\because {\sin }^{2}x\le 1\Rightarrow \frac{{\sin }^{2}x}{x^3}\le \frac{1}{x^3}\right)}\\ \\ ={\left[-\frac{1}{2x^2}\right]}_{10}^{\infty }\\ \\ =\frac{1}{200}\\ \\ =0.005\\ \\ \therefore R_n\le 0.005\end{array}$

Thus, the estimated of the given series is $0.83253$ and error is less than $0.005$ .