Problem 37
Question
Determine whether the series converges conditionally or absolutely, or diverges. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) !} $$
Step-by-Step Solution
Verified Answer
The series both conditionally and absolutely converges.
1Step 1: Identify the Series Test to Use
The series is an alternating series because of the \((-1)^n\) term. We will also need to consider the Absolute Convergence Test, so we will use both the Alternating Series Test and the Absolute Convergence Test.
2Step 2: Apply the Alternating Series Test
For the Alternating Series Test, the two conditions that need to be satisfied are: 1) The limit of the sequence must be zero and 2) The sequence must be decreasing. In this case, we have \(\lim_{n \to \infty} \frac{1}{(2n+1)!} = 0\) and \(\frac{1}{(2n+1)!} > \frac{1}{(2(n+1)+1)!}\) for all \(n\), so both conditions are satisfied, implying that the series converges conditionally.
3Step 3: Apply the Absolute Convergence Test
To apply the Absolute Convergence Test, we take the absolute value of the terms in the series, getting rid of the \(-1^n\) to have the series \(\sum_{n=0}^{\infty} \frac{1}{(2n+1)!}\). Using the Ratio Test on this series, we see that \(\frac{\frac{1}{(2(n+1)+1)!}}{\frac{1}{(2n+1)!}} = \frac{(2n+1)!}{(2n+1+2)!} = 0 < 1\), so this series absolutely converges.
Key Concepts
Alternating Series TestAbsolute Convergence TestRatio Test
Alternating Series Test
The Alternating Series Test is a useful tool for determining the convergence of series that alternate in sign. This test applies specifically to series that have the form
For the Alternating Series Test to conclude convergence, two conditions must be met:
- \( \sum_{n=0}^{\infty} (-1)^n a_n \)
For the Alternating Series Test to conclude convergence, two conditions must be met:
- **The limit condition:** The limit of the sequence \( a_n \) as \( n \) approaches infinity must be zero: \[ \lim_{n \to \infty} a_n = 0. \]
- **The decreasing condition:** The sequence \( a_n \) must be decreasing, meaning each term is smaller than the previous term. Thus, \[ a_{n+1} \leq a_n \text{ for all } n. \]
Absolute Convergence Test
When investigating series, we often need to test for absolute convergence to understand the behavior of the series more comprehensively. A series \( \sum_{n=0}^{\infty} a_n \) is said to converge absolutely if the series formed by taking the absolute values of its terms also converges:
To check for absolute convergence, we compare the absolute series to a known converging series or use a specific test like the Ratio Test, which is often quite effective.
- \[ \sum_{n=0}^{\infty} |a_n| \]
To check for absolute convergence, we compare the absolute series to a known converging series or use a specific test like the Ratio Test, which is often quite effective.
Ratio Test
The Ratio Test is a powerful tool for determining the convergence of a series. It involves looking at the ratio of consecutive terms in the series and is often used when dealing with factorials, exponential terms, or other complex expressions. To apply the Ratio Test, we examine the series: \( \sum_{n=0}^{\infty} a_n \).
The Ratio Test involves calculating the limit:
The Ratio Test involves calculating the limit:
- \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \]
- If \( L < 1 \), the series converges absolutely.
- If \( L > 1 \) or the limit does not exist, the series diverges.
- If \( L = 1 \), the ratio test is inconclusive, and another method must be used.
Other exercises in this chapter
Problem 37
In Exercises \(35-38,\) write an equivalent series with the index of summation beginning at \(n=1\). $$ \sum_{n=0}^{\infty} \frac{x^{2 n+1}}{(2 n+1) !} $$
View solution Problem 37
In Exercises \(35-38,\) use Taylor's Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error. $$ \arcsi
View solution Problem 37
Explain how to use the geometric series $$g(x)=\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}, \quad|x|
View solution Problem 38
Find the sum of the convergent series. $$ \sum_{n=0}^{\infty} 2\left(-\frac{2}{3}\right)^{n} $$
View solution