Answer
$n=8$
Work Step by Step
We have ${a_k} = \dfrac{1}{{k!}}$.
According to Theorem 9.6.2, the absolute error $\left| {S - {s_n}} \right|$ is given by Eq. (4):
$\left| {S - {s_n}} \right| \le {a_{n + 1}}$
For an accuracy to $0.00001$, we must have
${a_{n + 1}} \le 0.00001$
$\dfrac{1}{{\left( {n + 1} \right)!}} \le \dfrac{1}{{100000}}$
$\left( {n + 1} \right)! \ge 100000$
Using a calculating utility, we solve this inequality and obtain $n \ge 7.42$. Thus, the closest integer is $n=8$.
