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.