Approximate $e$ correct to within 0.000001.

Solution In Example 2 (p. 615) of Section 8.9, we showed that the Maclaurin series $\sum\limits_{k\,=\,0}^{\infty }\dfrac{x^{k}}{k!}$ converges to $e^{x}$ for every number $x.$ If $x=1,$ we have $$e=e^{1}=1+1+\dfrac{1}{2!}+\dfrac{1}{3!}+\cdots \,+\dfrac{1}{n!}+\cdots\, =\sum\limits_{k\,=\,0}^{\infty }\dfrac{1}{k!}$$

But this series converges very slowly.

For faster convergence, we can use $x=-1.$ Then $$e^{-1}=1-1+\dfrac{1}{2!}-\dfrac{1}{3!}+\cdots \,+\dfrac{(-1) ^{n}}{n!}+\cdots \,=\sum\limits_{k\,=\,0}^{\infty }\dfrac{(-1) ^{k}}{k!}$$

Since this series is an alternating series that satisfies the conditions of the Alternating Series Test, we can use the error estimate for an alternating series. Since we want $e$ correct to within 0.000001, we need the error $E<0.000001.$ That is, we need $\dfrac{1}{n!}<0.000001,$ or equivalently, $n!>10^{6}$. Since $10!= 3.6288\times 10^{6}>10^{6},$ but $9!= 362{,}880<10^{6},$ using $10$ terms of the Maclaurin series will approximate $e^{-1}$ correct to within 0.000001.

626

$$\sum\limits_{k\,=\,0}^{9}\dfrac{(-1) ^{k}}{k!}=1-1+\dfrac{1}{2}- \dfrac{1}{3!}+\dfrac{1}{4!}-\dfrac{1}{5!}+\dfrac{1}{6!}-\dfrac{1}{7!}+\dfrac{ 1}{8!}-\dfrac{1}{9!}=\dfrac{16{,}687}{45{,}360}$$

Then $$e=\dfrac{1}{e^{-1}}= \left( \dfrac{16{,}687}{45{,}360}\right)^{-1}\approx 2.\, 718\,28$$

correct to within 0.000001.