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Chapter 9: Problem 102

Use a graphing utility to find the sum. $$\sum_{k=0}^{4} \frac{(-1)^{k}}{k !}$$

Short Answer

Expert verified
The sum of the series is 0.375.

Step by step solution

01

Understand the summation

The notation \( \sum_{k=0}^{4} \frac{(-1)^{k}}{k!} \) means that we are summing up a series from k=0 to k=4, where the series is defined as \(\frac{(-1)^{k}}{k!}\). In this series, the term \((-1)^k\) alternates in sign with each increase in k, and the term \(\frac{1}{k!}\) includes the factorial of k in the denominator. A factorial is the product of all positive integers less than or equal to that integer.
02

Calculate each term

With the understanding of what each part of the series represents, we can then calculate each individual term in the series from k=0 to k=4: When k=0, the term would be \(\frac{(-1)^0}{0!} = 1\). When k=1, the term would be \(\frac{(-1)^1}{1!} = -1\). When k=2, the term would be \(\frac{(-1)^2}{2!} = \frac{1}{2}\). When k=3, the term would be \(\frac{(-1)^3}{3!} = -\frac{1}{6}\). And when k=4, the term would be \(\frac{(-1)^4}{4!} = \frac{1}{24}\).
03

Sum up the terms

We then sum up all individual terms that we calculated in the previous step. The overall sum of the series is \(1 - 1 + 0.5 - \frac{1}{6} + \frac{1}{24} = 0.375\).

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Most popular questions from this chapter

$$1+\frac{1}{1}, 1+\frac{1}{2}, 1+\frac{1}{3}, 1+\frac{1}{4}, 1+\frac{1}{5}, \dots$$$$1+\frac{1}{1}, 1+\frac{1}{2}, 1+\frac{1}{3}, 1+\frac{1}{4}, 1+\frac{1}{5}, \dots$$

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