Solutions - 0250-01

Notice that ${\displaystyle \frac{1}{e}}=e^{-1}$.

We have this series for $e^x$:

$$e^x=\sum _{n=0}^{\infty }\frac{x^n}{n!}$$

Therefore:

$$\begin{array}{c}{e}^{-1}=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{n!}=\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}\\ \\ \gg \gg \frac{1}{e}=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\cdots \end{array}$$

Now we use the “Next Term Bound” rule. Calculate terms until we find a term less than ${10}^{-3}$:

$$\begin{array}{c}\frac{1}{4!}=4.2\times {10}^{-2},\frac{1}{5!}=8.3\times {10}^{-3},\\ \\ \frac{1}{6!}=1.4\times {10}^{-3},\frac{1}{7!}=2.4\times {10}^{-4}\end{array}$$

So we take the following partial sum approximation:

$$\frac{1}{e}\approx \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}=0.3681$$