## Calculus 8th Edition

Published by Cengage

# Chapter 11 - Infinite Sequences and Series - 11.1 Sequences - 11.1 Exercises - Page 744: 7

#### Answer

$\frac{1}{2},\frac{1}{6},\frac{1}{24},\frac{1}{120},\frac{1}{720}$

#### Work Step by Step

To find the first five terms of the sequence $a_n = \frac{1}{(n+1)!}$, we must plug in $n=1, n=2, n=3, n=4,$ and $n=5.$ $a_1 = \frac{1}{(1+1)!}=\frac{1}{2!}=\frac{1}{2}$ $a_2 =\frac{1}{(2+1)!}=\frac{1}{3!}=\frac{1}{6}$ $a_3 =\frac{1}{(3+1)!}=\frac{1}{4!}=\frac{1}{24}$ $a_4 = \frac{1}{(4+1)!}=\frac{1}{5!}=\frac{1}{120}$ $a_5 = \frac{1}{(5+1)!}=\frac{1}{6!}=\frac{1}{720}$ Hence we see that the first five terms are $\frac{1}{2},\frac{1}{6},\frac{1}{24},\frac{1}{120},\frac{1}{720}$

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