Chapter 8, Problem 26RQ

(a)

To determine

To Find: The value of the $s_5$ of the series ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^6}}$ and then estimate the error in using it as the approximation to sum of the series.

Answer to Problem 26RQ

The size of the error is $6.4\times {10}^{-5}$ .

Explanation of Solution

Given:

The given expression is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^6}}$

Calculation:

Consider the given series is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^6}}$

Then, the sum is,

  $\begin{array}{c}{s}_5=\frac{1}{1^6}+\frac{1}{2^6}+\frac{1}{3^6}+\frac{1}{4^6}+\frac{1}{5^6}\\ =1.017305\end{array}$

Then, by the remainder estimate solve as,

  $\begin{array}{c}{R}_5\le {\displaystyle {\int }_5^{\infty }\frac{1}{x^6}dx}=\underset{t\to \infty }{\lim }{\left[-\frac{1}{5x^5}\right]}_5^t\\ R_5\le \left[0+\frac{1}{5^6}\right]=6.4\times {10}^{-5}\end{array}$

Thus, the size of the error is $6.4\times {10}^{-5}$ .

(b)

To determine

To Find: The sum of the series that is correct to live decimal places.

Answer to Problem 26RQ

The sum of the series is $1.01734$ .

Explanation of Solution

Calculation:

To obtain the sum correct to five decimal places that must have the error size ${10}^{-6}$ , then $R_n\le {10}^{-6}$ and the obtain the number of terms by the use of remainder estimate is,

  $R_n\le {\displaystyle {\int }_n^{\infty }\frac{1}{x^6}dx}=\frac{1}{5n^5}$

Solving the above inequalities as,

  $\begin{array}{l}5n^5>\frac{1}{0.00001}\\ n>11.48\end{array}$

Then, the sum of the 12 terns is,

  $\begin{array}{c}{s}_{12}=\frac{1}{1^6}+\frac{1}{2^6}+\frac{1}{3^6}+\frac{1}{4^6}+\frac{1}{5^6}+\frac{1}{6^6}+\mathrm{.....}+\frac{1}{{12}^6}\\ =1.01734\end{array}$