Chapter 6.6, Problem 1CE

To determine

Calculate the value of the each given series.

Answer to Problem 1CE

The solution is

  $\begin{array}{l}1+\frac{1}{1!}+\frac{1}{2!}=2.5\\ 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}=2.667\\ 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}=2.7083\\ 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}=2.7167\\ 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}=2.7181\\ 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!}=2.7183\end{array}$

Explanation of Solution

Given: The series are given

  $\begin{array}{l}1+\frac{1}{1!}+\frac{1}{2!}\\ 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}\\ 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}\\ 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}\\ 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}\\ 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!}\end{array}$

Calculation:

Using MATLAB, we have calculated the value of series

Let’s take one series

  $1+\frac{1}{1!}+\frac{1}{2!}$

  $1+\frac{1}{1!}+\frac{1}{2!}=1+1+0.5=2.5$

program:

clc
clear
close all
n=7;
for i=1:1:n
    a(i)=1./factorial(i);
    a1=1+sum(a);
end

Query:

• First, we have defined the value of the n.
• Then using simple relation as given to find the solution.