52613912
Euler's number can be represented by the infinite series \(e=\sum_{k=0}^{\infty}\frac{1}{k!}=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots\). Define the partial sums by \(s_n=\sum_{k=0}^{n}\frac{1}{k!}\).
1. Calculate \(s_5\). Round the result to four decimal places.
2. The error after stopping the series at \(\frac{1}{n!}\) is \(R_n=e-s_n\), and it satisfies \(R_n<\frac{1}{n\cdot n!}\). Use this bound to find the smallest value of \(n\) for which the error is guaranteed to be less than \(10^{-4}\).
3. Compare the efficiency of this series with the sequence \(a_n=\left(1+\frac{1}{n}\right)^n\), which also approaches \(e\). Briefly explain which method is better for obtaining an accurate approximation of \(e\).
Hints
- Recall that \(n!\) is the product of the positive integers from \(1\) through \(n\).
- Test consecutive small positive integers in the error bound.
- Compare how quickly the denominators or approximation errors change as \(n\) increases.
Solution
1. \(s_5=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}=\frac{163}{60}\approx2.7167\).
2. We need \(\frac{1}{n\cdot n!}<10^{-4}\), or equivalently \(n\cdot n!>10{,}000\). For \(n=6\), \(6\cdot6!=4320<10{,}000\). For \(n=7\), \(7\cdot7!=35{,}280>10{,}000\). Therefore, the smallest value is \(n=7\).
3. The factorial in the denominator of the series terms grows very quickly, so the partial sums approach \(e\) rapidly. The sequence \(\left(1+\frac{1}{n}\right)^n\) approaches \(e\) much more slowly. The series is therefore more efficient for high-accuracy approximation.
Answer
1. \(s_5\approx2.7167\)
2. \(n=7\)
3. The series method is more efficient because its error decreases much faster.
