Solutions - 0190-03

We use the alternating series test error bound formula. (AKA: “Next Term Bound”)

$$|E_n|\le a_{n+1}$$

We seek the smallest $n$ such that $a_{n+1}<0.005$. What that happens, we will have:

$$|E_n|\le a_{n+1}<0.005\gg \gg |E_n|<0.005$$

Our formula for $a_n$:

$$a_n=\frac{1}{n!}\gg \gg a_{n+1}=\frac{1}{(n+1)!}$$

We cannot easily solve for $n$ to provide $a_{n+1}<0.005$, so we just start listing out the terms:

$$\begin{array}{c}{a}_1=1,a_2=0.5,a_3=0.167,\\ \\ a_4=0.0417,a_5=0.00833,a_6=0.00139\end{array}$$

We see that $a_6$ is the first term less than 0.005, so $n+1=6$ and we need the first 5 terms:

$$\sum _{n=1}^5\frac{{(-1)}^{n-1}}{n!}=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\frac{1}{5!}\gg \gg \approx 0.633$$