Chapter 8.3, Problem 37E

To determine

To find: The sum of first 10 terms and estimate error.

Answer to Problem 37E

  $s_{10}\approx 1.2486$ with remainder $R_{10}\le 0.1$ .

Explanation of Solution

Given information: Given series ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n^4+1}}}$

${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n^4+1}}}=a_n$

Sum of the first 10 terms

  $\begin{array}{l}{s}_{10}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{17}}+\frac{1}{\sqrt{82}}+\mathrm{......}+\frac{1}{\sqrt{10001}}\\ \therefore s_{10}=1.2486\end{array}$

Also

  $\begin{array}{l}{b}_n=\frac{1}{\sqrt{n^4}}=\frac{1}{n^2}\\ \Rightarrow b_n=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\mathrm{......}\end{array}$

Here, $\begin{array}{l}{b}_n>a_n\\ \therefore a_n\text{ converges}\end{array}$

Now from the error(remainder) calculation,

  $\begin{array}{l}{R}_{10}\le {\displaystyle \underset{10}{\overset{\infty }{\int }}\frac{1}{X^2}}dX\\ \Rightarrow R_{10}\le {\left[\frac{X^{-2+1}}{-2+1}\right]}_{10}^{\infty }\\ \Rightarrow R_{10}\le {\left[-\frac{1}{X}\right]}_{10}^{\infty }\\ \Rightarrow R_{10}\le \left[-(\frac{1}{\infty }-\frac{1}{10})\right]\\ \Rightarrow R_{10}\le \left[-(0-\frac{1}{10})\right]\\ \Rightarrow R_{10}\le \frac{1}{10}\\ \Rightarrow R_{10}\le 0.1\end{array}$

So, $s_{10}\approx 1.2486$ with remainder $R_{10}\le 0.1$