Answer
The approximated sum is ${s_3} \approx 0.842$.
Work Step by Step
Write
$S = 1 - \dfrac{1}{{3!}} + \dfrac{1}{{5!}} - \dfrac{1}{{7!}} + \cdot\cdot\cdot = \mathop \sum \limits_{k = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{k - 1}}}}{{\left( {2k - 1} \right)!}}$,
where ${a_k} = \dfrac{1}{{\left( {2k - 1} \right)!}}$.
According to Theorem 9.6.2, the upper bound on the absolute error is ${a_{n + 1}}$, given by Eq. (4):
$\left| {S - {s_n}} \right| \le {a_{n + 1}}$
Thus, for two decimal-place accuracy, we choose a value of $n$ such that
$\left| {S - {s_n}} \right| \le {a_{n + 1}} \lt 0.005$
$\dfrac{1}{{\left( {2n + 1} \right)!}} \lt 0.005$
$\left( {2n + 1} \right)! \gt 200$
Solving this inequality using a calculating utility, we obtain $n \gt 2.15$. Thus, we can take $n=3$.
Thus, the approximated sum is ${s_3}$:
${s_3} = 1 - \dfrac{1}{{3!}} + \dfrac{1}{{5!}} = \dfrac{{101}}{{120}} \approx 0.842$
