Chapter 8.3, Problem 33E

a.

To determine

To estimate: The sum of the given series using the sum of the first $10$ terms.

Answer to Problem 33E

The estimated sum of the given series is $1.549768$ with error less than $0.1$ .

Explanation of Solution

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

Calculation:

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

  $s_{10}=a_1+a_2+\mathrm{....}+a_{10}=1+\frac{1}{2^2}+\mathrm{....}+\frac{1}{{10}^2}\approx 1.549768$

According to the Remainder Estimate,

  $\begin{array}{l}{R}_n=s-s_n\le {\displaystyle \underset{n}{\overset{\infty }{\int }}f(x)dx}\\ \\ \Rightarrow R_{10}=s-s_{10}\le {\displaystyle \underset{10}{\overset{\infty }{\int }}\frac{1}{x^2}dx}\\ \\ ={\left[-\frac{1}{x}\right]}_{10}^{\infty }\\ \\ =0+\frac{1}{10}\\ \\ =0.1\\ \\ \Rightarrow R_n\le 0.1\end{array}$

Thus, the estimated sum of the given series is $1.549768$ with error less than $0.1$ .

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 33E

The improved estimated sum of the given series is $1.645222$ .

Explanation of Solution

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

Calculation:

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

  $\begin{array}{l}{s}_{10}+{\displaystyle \underset{11}{\overset{\infty }{\int }}\frac{1}{x^2}dx\le s\le s_{10}}+{\displaystyle \underset{10}{\overset{\infty }{\int }}\frac{1}{x^2}dx}\\ \\ \Rightarrow 1.549768+{\left[-\frac{1}{x}\right]}_{11}^{\infty }\le s\le 1.549768+0.1(\because s_{10}=1.549768)\\ \\ \Rightarrow 1.549768+\frac{1}{11}\le s\le 1.649768\\ \\ \Rightarrow 1.640677\le s\le 1.649768\end{array}$

  $s=\frac{1.640677+1.649768}{2}=1.645222$

Thus, the improved estimated sum of the given series is $1.645222$ .

c.

To determine

To find: A value of $n$ that ensures that the error in the approximation $s\approx s_n$ is less than $0.001$ .

Answer to Problem 33E

  $n\ge 1000$ ensures that the error in the approximation $s\approx s_n$ is less than $0.001$ .

Explanation of Solution

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

Calculation:

Here, $f(x)=\frac{1}{x^2}$

  $\begin{array}{l}\Rightarrow R_n\le {\displaystyle \underset{n}{\overset{\infty }{\int }}\frac{1}{x^2}dx}\\ \\ ={\left[-\frac{1}{x}\right]}_n^{\infty }\\ \\ =\frac{1}{n}\end{array}$

So, $\frac{1}{n}\le 0.001$

  $\Rightarrow n\ge 1000$

Thus, $n\ge 1000$ ensures that the error in the approximation $s\approx s_n$ is less than $0.001$ .