## University Calculus: Early Transcendentals (3rd Edition)

$0.3678819444$
We have $S_n=a_1+a_2+......+(-1)^{n+1}a_n$ Now, $|S-S_n| \leq |a_{n+1}| \implies \space |Error| \lt |a_{n+1} |$ or, $|\dfrac{1}{k!}| \lt \dfrac{5}{10^6}$ or, $k \ge 9$ Thus, we need at least 9 terms. So, $S \approx S_8 =1-1+\dfrac{1}{2!}-\dfrac{1}{3!}+.... \approx 0.3678819444$