Chapter 6.6, Problem 2CE

To determine

Calculate the value of the each given series.

Answer to Problem 2CE

The solution is

  $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\mathrm{......}+\frac{1}{50!}=2.7183$

Explanation of Solution

Given: The series are given

  $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\mathrm{......}+\frac{1}{n!}$

Calculation:

Using MATLAB, we have calculated the value of series

Let’s take n=50 one series

  $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\mathrm{......}+\frac{1}{50!}$

  $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\mathrm{......}+\frac{1}{50!}=2.7183$

So, as we were seen that in previous problem the value of the series 7! Is same as the above

  $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!}=2.7183$

So, we can say that after 7! the value of the series is constant.

program:

clc
clear
close all
n=50;
fori=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.