Chapter 8.3, Problem 32E

a.

To determine

To find: The partial sum $s_{10}$ of the given series and to estimate the error in using $s_{10}$ as an approximation to the sum of the series.

Answer to Problem 32E

  $s_{10}=1.082037$ and the error in using $s_{10}$ as an approximation to the sum of the series is

  $0.000\overline{3}$ .

Explanation of Solution

Given Information: Given series is ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^4}}$ .

Calculation:

Sum of first $10$ terms of the given series is:

  $s_{10}=a_1+a_2+\mathrm{....}+a_{10}=1+\frac{1}{2^4}+\mathrm{....}+\frac{1}{{10}^4}\approx 1.082037$

According to the Remainder Estimate,

  $\begin{array}{l}{R}_n=s-s_n\le {\displaystyle \underset{n}{\overset{\infty }{\int }}f(x)dx}\\ \\ \Rightarrow R_{10}=s-s_{10}\le {\displaystyle \underset{10}{\overset{\infty }{\int }}\frac{1}{x^4}dx}\\ \\ ={\left[-\frac{1}{3x^3}\right]}_{10}^{\infty }\\ \\ =0+\frac{1}{3000}\\ \\ =0.000\overline{3}\\ \\ \Rightarrow R_n\le 0.000\overline{3}\end{array}$

Thus, $s_{10}=1.082037$ and the error in using $s_{10}$ as an approximation to the sum of the series is

  $0.000\overline{3}$ .

b.

To determine

To improve: The estimate in part (a) using the inequality $s_n+{\displaystyle \underset{n+1}{\overset{\infty }{\int }}f(x)dx\le s\le s_n}+{\displaystyle \underset{n}{\overset{\infty }{\int }}f(x)dx}$ with $n=10$ .

Answer to Problem 32E

The improved estimated sum of the given series is $1.082328$ .

Explanation of Solution

Given Information: Given series is ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^4}}$ .

Calculation:

Putting $n=10$ in $s_n+{\displaystyle \underset{n+1}{\overset{\infty }{\int }}f(x)dx\le s\le s_n}+{\displaystyle \underset{n}{\overset{\infty }{\int }}f(x)dx}$ ,

  $\begin{array}{l}{s}_{10}+{\displaystyle \underset{11}{\overset{\infty }{\int }}\frac{1}{x^4}dx\le s\le s_{10}}+{\displaystyle \underset{10}{\overset{\infty }{\int }}\frac{1}{x^4}dx}\\ \\ \Rightarrow 1.082037+{\left[-\frac{1}{3x^3}\right]}_{11}^{\infty }\le s\le 1.082037+{\left[-\frac{1}{3x^3}\right]}_{10}^{\infty }(\because s_{10}=1.082037)\\ \\ \Rightarrow 1.549768+0.00025\le s\le 1.082037+0.00033\\ \\ \Rightarrow 1.082287\le s\le 1.082370\end{array}$

  $s=\frac{1.082370+1.082287}{2}=1.082328$

Thus, the improved estimated sum of the given series is $1.082328$ .

c.

To determine

To find: A value of $n$ so that $s_n$ is within $0.00001$ of the sum.

Answer to Problem 32E

  $n\ge 33$ so that $s_n$ is within $0.00001$ of the sum.

Explanation of Solution

Given Information: Given series is ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^4}}$ .

Calculation:

Here, $f(x)=\frac{1}{x^4}$

  $\begin{array}{l}\Rightarrow R_n\le {\displaystyle \underset{n}{\overset{\infty }{\int }}\frac{1}{x^4}dx}\\ \\ ={\left[-\frac{1}{3x^3}\right]}_n^{\infty }\\ \\ =\frac{1}{3n^3}\end{array}$

So, $\frac{1}{3n^3}\le 0.00001$

  $\Rightarrow n^3\ge 33333.33$

  $\Rightarrow n\ge 32.18$

Thus, $n\ge 33$ so that $s_n$ is within $0.00001$ of the sum.