Problem 47
Question
Which of the series Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \frac{1}{4}-\frac{1}{6}+\frac{1}{8}-\frac{1}{10}+\frac{1}{12}-\frac{1}{14}+\cdots $$
Step-by-Step Solution
Verified Answer
The series converges, but not absolutely.
1Step 1: Identify the Series Type
The given series is an alternating series because it has terms that alternate between positive and negative signs: \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{4+2n} \). This resembles an alternating harmonic series.
2Step 2: Apply the Alternating Series Test
For an alternating series \( \sum (-1)^n a_n \), if the terms \( a_n \) are positive, decreasing, and approach zero as \( n \to \infty \), then the series converges. Here, \( a_n = \frac{1}{4+2n} \), which is positive, decreases, and approaches zero as \( n \to \infty \). Thus, the series converges.
3Step 3: Check for Absolute Convergence
Test for absolute convergence by considering the series without the alternating sign: \( \sum_{n=1}^{\infty} \frac{1}{4+2n} \). This behaves like a harmonic series \( \sum \frac{1}{n} \), which is known to diverge because the harmonic series diverges.
4Step 4: Conclusion
The original series converges by the Alternating Series Test, but it does not converge absolutely because the series of absolute values diverges.
Key Concepts
Convergence testsAbsolute convergenceDivergent series
Convergence tests
When it comes to determining whether a series converges, convergence tests are invaluable tools. They help us decide if adding up an infinite series results in a finite value. For alternating series, like the one in the original exercise, the Alternating Series Test is a popular choice.
**Here's how this test works:**
* It analyzes series where the terms switch signs, such as from positive to negative.
* To apply this test, each term (without considering the sign) must be positive, decrease as the series progresses, and approach zero as the number of terms goes to infinity.
**Here's how this test works:**
* It analyzes series where the terms switch signs, such as from positive to negative.
* To apply this test, each term (without considering the sign) must be positive, decrease as the series progresses, and approach zero as the number of terms goes to infinity.
- If these criteria are met, the series converges.
Absolute convergence
Absolute convergence is a stronger form of convergence. A series converges absolutely if the series formed by taking the absolute value of each term also converges. This concept is essential because absolute convergence implies regular convergence, but not vice versa.
To check for absolute convergence, we first "ignore" the signs in the series. In the original problem, we looked at the series without alternating signs: \( \sum_{n=1}^{\infty} \frac{1}{4+2n} \). Unfortunately, this resembles the harmonic series \( \sum \frac{1}{n} \), which we know diverges.
To check for absolute convergence, we first "ignore" the signs in the series. In the original problem, we looked at the series without alternating signs: \( \sum_{n=1}^{\infty} \frac{1}{4+2n} \). Unfortunately, this resembles the harmonic series \( \sum \frac{1}{n} \), which we know diverges.
- This means the absolute series does not converge.
Divergent series
A series is said to diverge when the sum of its terms doesn't settle on a particular numerical value, even if the series is extended indefinitely. Recognizing a divergent series is vital because it tells us about the behavior of the series when attempting to sum an endless number of terms.
The absolute series in the exercise is a prime example. Without alternating signs, it behaves like a harmonic series. In mathematics, the harmonic series \( \sum \frac{1}{n} \) famously diverges as its terms don't approach a fixed sum, even though they grow smaller.
The absolute series in the exercise is a prime example. Without alternating signs, it behaves like a harmonic series. In mathematics, the harmonic series \( \sum \frac{1}{n} \) famously diverges as its terms don't approach a fixed sum, even though they grow smaller.
- Since it diverges, it highlights that our series cannot converge absolutely.
Other exercises in this chapter
Problem 47
The (second) second derivative test Use the equation $$f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}\left(c_{2}\right)}{2}(x-a)^{2}$$ to establish the fo
View solution Problem 47
Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{\tan ^{-1} n}{n^{1
View solution Problem 47
Find the sum of each series in Exercises \(41-48\) $$ \sum_{n=1}^{\infty}\left(\frac{1}{\ln (n+2)}-\frac{1}{\ln (n+1)}\right) $$
View solution Problem 47
Recursively Defined Terms Which of the series \(\sum_{n=1}^{\infty} a_{n}\) defined by the formulas in Exercises \(45-54\) converge, and which diverge? Give rea
View solution