Chapter 5.3, Problem 116AYU

To determine

To find: Use a calculator to compute the values of.

  $2+\frac{1}{2!}+\frac{1}{3!}+\dots +\frac{1}{n!}$

for n = 4, 6, 8, and 10.

Compare each result with e.

Answer to Problem 116AYU

Solution:

  $e=2+ \frac{1}{2!}+ \frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!}+\frac{1}{8!}+\frac{1}{9!}+\frac{1}{10!}$

Explanation of Solution

Given:

  $2+ \frac{1}{2!}+ \frac{1}{3!}+ \dots +\frac{1}{n!}$

Calculation:

We know that $e = 2.7183$ -------------(1)

Given $2+ \frac{1}{2!}+ \frac{1}{3!}+ \dots +\frac{1}{n!}$

When $n=4$

  $2+ \frac{1}{2!}+ \frac{1}{3!}+\frac{1}{4!}=2.7083$ -------------(2)

Compare (1) and (2) we get

  $e>2+ \frac{1}{2!}+ \frac{1}{3!}+\frac{1}{4!}$

When $n=6$

  $2+ \frac{1}{2!}+ \frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}=2.7181$ -------------(3)

Compare (1) and (3) we get

  $e>2+ \frac{1}{2!}+ \frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}$

When $n=8$

  $2+ \frac{1}{2!}+ \frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!}+\frac{1}{8!}=2.7183$ -------------(4)

Compare (1) and (4) we get

  $e=2+ \frac{1}{2!}+ \frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!}+\frac{1}{8!}$

When $n=10$

  $2+ \frac{1}{2!}+ \frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!}+\frac{1}{8!}+\frac{1}{9!}+\frac{1}{10!}=2.7183$ -------------(5)

Compare (1) and (5) we get

  $e=2+ \frac{1}{2!}+ \frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!}+\frac{1}{8!}+\frac{1}{9!}+\frac{1}{10!}$