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Chapter 7: Problem 47

Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{1}{n !} $$

Short Answer

Expert verified
The series \(\sum_{n=0}^{\infty} \frac{1}{n !}\) converges.

Step by step solution

01

Identify Known Series

In order to apply the Direct Comparison Test, a known series needs to be identified that can be compared to the given series. An applicable known series is the geometric series of \(\sum_{n=0}^{\infty} \frac{1}{2^n}\), which converges because the common ratio (1/2) is between -1 and 1.
02

Comparison

Now, compare the given series \(\sum_{n=0}^{\infty} \frac{1}{n !}\) with the known series \(\sum_{n=0}^{\infty} \frac{1}{2^n}\). For \(n\) greater than zero, \(\frac{1}{n !} <= \frac{1}{2^n}\), since \(n !\) is always less than or equal to \(2^n\). Therefore, the term \(a_n = \frac{1}{n !}\) of the given series is less than or equal to that of the known series \(b_n = \frac{1}{2^n}\).
03

Apply Direct Comparison Test

The Direct Comparison Test states that if \(0 <= a_n <= b_n\) for all \(n\) and \(\sum_{n=0}^{\infty} b_n\) converges, then \(\sum_{n=0}^{\infty} a_n\) also converges. Since \(\sum_{n=0}^{\infty} \frac{1}{2^n}\) is a convergent geometric series and \(0 <= \frac{1}{n !} <= \frac{1}{2^n}\), \(\sum_{n=0}^{\infty} \frac{1}{n !}\) also converges by the Direct Comparison Test.

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Most popular questions from this chapter

Use a graphing utility to determine the first term that is less than 0.0001 in each of the convergent series. Note that the answers are very different. Explain how this will affect the rate at which each series converges. $$ \sum_{n=1}^{\infty} \frac{1}{2^{n}}, \quad \sum_{n=1}^{\infty}(0.01)^{n} $$

In Exercises 87 and 88 , use a graphing utility to graph the function. Identify the horizontal asymptote of the graph and determine its relationship to the sum of the series. $$ \frac{\text { Function }}{f(x)=3\left[\frac{1-(0.5)^{x}}{1-0.5}\right]} \frac{\text { Series }}{\sum_{n=0}^{\infty} 3\left(\frac{1}{2}\right)^{n}} $$

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{2^{n}}{100} $$

(a) write the repeating decimal as a geometric series and (b) write its sum as the ratio of two integers $$ 0.0 \overline{75} $$

The Fibonacci sequence is defined recursively by \(a_{n+2}=a_{n}+a_{n+1},\) where \(a_{1}=1\) and \(a_{2}=1\) (a) Show that \(\frac{1}{a_{n+1} a_{n+3}}=\frac{1}{a_{n+1} a_{n+2}}-\frac{1}{a_{n+2} a_{n+3}}\). (b) Show that \(\sum_{n=0}^{\infty} \frac{1}{a_{n+1} a_{n+3}}=1\).
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