Chapter 8, Problem 27RQ

To determine

To Find: The sum of the first eight terms of the approximate the sum of the series that is ${\displaystyle \sum _{n=1}^{\infty }{\left(2+5^n\right)}^{-1}}$ .

Answer to Problem 27RQ

The sum of the first eight term is $6.4\times {10}^{-7}$ .

Explanation of Solution

Given:

The given expression is,

  ${\displaystyle \sum _{n=1}^{\infty }{\left(2+5^n\right)}^{-1}}$

Calculation:

Consider the given series is,

  ${\displaystyle \sum _{n=1}^{\infty }{\left(2+5^n\right)}^{-1}}$

Then, the sum is,

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }{\left(2+5^n\right)}^{-1}}={\displaystyle \sum _{n=1}^8{\left(2+5^n\right)}^{-1}}\\ =\frac{1}{2+5}+\frac{1}{2+5^2}+\frac{1}{2+5^3}+\frac{1}{2+5^4}+\mathrm{...}+\frac{1}{2+5^8}\\ =0.18976224\end{array}$

Consider the case $\frac{1}{2+5^n}<\frac{1}{5^n}$ for all $n$ . Then, the remainder term is,

  $R_8={\displaystyle \sum _{n=9}^{\infty }\frac{1}{2+5^n}}<{\displaystyle \sum _{n=9}^{\infty }\frac{1}{5^n}}$

As, ${\displaystyle \sum _{n=9}^{\infty }\frac{1}{5^n}}$ is the geometric series is with the $a=\frac{1}{5^9}$ and $r=\frac{1}{n}$ .

Then,

  $R_8<\frac{\frac{1}{5^9}}{1-\frac{1}{5}}=6.4\times {10}^{-7}$