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Magazine Chapter 9: Problem 1
Determine whether each series converges absolutely, converges conditionally, or diverges. $$\sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{2 n^{2}+3 n+5}$$
Short Answer
Expert verified
The series converges absolutely.
Step by step solution
01
Identify the Form of the Series
The given series \( \sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{2n^{2}+3n+5} \) is an alternating series due to the presence of the \((-1)^{n-1}\) term, which causes the signs of the terms to alternate.
02
Use the Alternating Series Test
The Alternating Series Test states that an alternating series \( \sum_{n=1}^{\infty} (-1)^{n-1} a_n \) converges if the sequence \( a_n \) is decreasing and \( \lim_{n\to\infty} a_n = 0 \). Here, \( a_n = \frac{1}{2n^{2} + 3n + 5} \).
03
Check Decreasing Condition
To verify that \( a_n \) is decreasing, note that the function \( f(n) = \frac{1}{2n^2 + 3n + 5} \) is positive and its denominator increases as \( n \) increases, making \( f(n) \) decrease for increasing \( n \). Thus, \( a_n \) is decreasing.
04
Check the Limit Condition
Calculate \( \lim_{n \to \infty} \frac{1}{2n^{2} + 3n + 5} \). As \( n \) approaches infinity, \( \frac{1}{2n^2 + 3n + 5} \) approaches 0 because the degree of the polynomial in the denominator (quadratic) remains constant as \( n \) increases.
05
Conclude the Alternating Series Test
Since \( a_n \) is decreasing and \( \lim_{n \to \infty} a_n = 0 \), the Alternating Series Test tells us that the series converges. Hence, the series converges.
06
Absolute Convergence Test
To determine if the series converges absolutely, consider the series \( \sum_{n=1}^{\infty} \left| (-1)^{n-1} \frac{1}{2n^{2}+3n+5} \right| = \sum_{n=1}^{\infty} \frac{1}{2n^{2} + 3n + 5} \). Use the comparison test.
07
Comparison with a Known Series
Note that \( \frac{1}{2n^{2} + 3n + 5} \) is bounded by \( \frac{1}{2n^{2}} \). The series \( \sum_{n=1}^{\infty} \frac{1}{2n^{2}} \) is a convergent p-series with \( p=2 \). Since the given series' terms are smaller, our series \( \sum_{n=1}^{\infty} \frac{1}{2n^{2} + 3n + 5} \) is convergent by the comparison test.
08
Conclusion on Absolute/Conditional Convergence
Since the series \( \sum_{n=1}^{\infty} \frac{1}{2n^{2} + 3n + 5} \) converges, the given alternating series converges absolutely. Therefore, the series converges absolutely.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Alternating Series Test
The Alternating Series Test (AST) is a method for determining the convergence of alternating series, which are characterized by terms that switch signs between positive and negative. The general form of an alternating series is: \[ \sum_{n=1}^{\infty} (-1)^{n-1} a_n,\]where each term alternates due to the factor \((-1)^{n-1}\). To apply this test, two conditions must be satisfied:
In our example, the sequence \(a_n = \frac{1}{2n^2 + 3n + 5}\) meets both conditions:
- The sequence \(a_n\) is decreasing, meaning each term is smaller than the previous one.
- The limit of \(a_n\) must approach zero as \(n\) tends to infinity.
In our example, the sequence \(a_n = \frac{1}{2n^2 + 3n + 5}\) meets both conditions:
- It decreases as \(n\) increases because the polynomial in the denominator grows, making each \(a_n\) smaller.
- The limit \( \lim_{n \to \infty} \frac{1}{2n^2 + 3n + 5} = 0\).
Absolute Convergence
Absolute convergence is a stronger form of convergence for series. A series is said to converge absolutely if the series of its absolute values also converges:\[\sum_{n=1}^{\infty} |a_n|.\]If a series converges absolutely, it definitively converges regardless of the order of its terms.
To determine absolute convergence for our series:
In conclusion, the presence of absolute convergence implies conditional convergence, but absolute convergence is a stronger statement ensuring the series remains convergent no matter how it's rearranged.
To determine absolute convergence for our series:
- Consider the series of absolute values:\[\sum_{n=1}^{\infty} \left| (-1)^{n-1}\frac{1}{2n^{2}+3n+5} \right| = \sum_{n=1}^{\infty} \frac{1}{2n^{2}+3n+5}.\]
- Now check if this new series converges using a comparison test.
In conclusion, the presence of absolute convergence implies conditional convergence, but absolute convergence is a stronger statement ensuring the series remains convergent no matter how it's rearranged.
Comparison Test
The Comparison Test is useful for showing convergence by comparing a series to another one whose convergence is known. This test is only applicable to series with positive terms.
Here's how it works:
Here's how it works:
- Given two series \( \sum a_n \) and \( \sum b_n \), to apply the test, we need to find a series \( \sum b_n \) such that \( 0 \leq a_n \leq b_n \) for all \(n\).
- If \( \sum b_n \) converges and \( a_n \leq b_n \), then \( \sum a_n \) also converges.
- If \( \sum b_n \) diverges and \( a_n \geq b_n \), then \( \sum a_n \) also diverges.
- Since \( \frac{1}{2n^{2}+3n+5} \leq \frac{1}{2n^{2}}\) for all \(n\), and the \(p\)-series converges, the comparison test shows that our series converges as well.
