Problem 27
Question
Determine whether the given series is convergent or divergent. $$ \sum_{n=1}^{\infty} \frac{1}{4 n^{2}-1} $$
Step-by-Step Solution
Verified Answer
The given series \(\sum_{n=1}^{\infty} \frac{1}{4n^2-1}\) is convergent by the Direct Comparison Test with the known convergent p-series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\).
1Step 1: Identify the known convergent series for comparison
We can choose the series:
$$
\sum_{n=1}^{\infty} \frac{1}{n^2}
$$
to make our comparison. This is a convergent p-series with p = 2 > 1. This implies that the series converges.
2Step 2: Direct Comparison Test
We want to compare the terms of the given series with the terms of the known convergent series:
$$
\frac{1}{4n^2 - 1} \text{ and } \frac{1}{n^2}
$$
We need to check if the given series' terms are smaller than or equal to the terms of the known convergent series. This inequality should hold for all n:
$$
\frac{1}{4n^2 - 1} \leq \frac{1}{n^2}
$$
3Step 3: Determine inequality
To find if the inequality is true, we can cross-multiply and simplify:
$$
n^2 (4n^2 - 1) \leq (4n^2 - 1)
$$
Simplifying, we have
$$
4n^4 - n^2 \leq 4n^2 - 1
$$
Rearranging the terms, we get
$$
4n^4 - 5n^2 +1 \geq 0
$$
Since the inequality holds for every term in the series and the right-hand side is a convergent p-series, we can conclude that our given series is also convergent by the Direct Comparison Test.
4Step 4: Final Step: Series Convergence
The given series:
$$
\sum_{n=1}^{\infty} \frac{1}{4 n^{2}-1}
$$
is convergent by the Direct Comparison Test with the known convergent p-series:
$$
\sum_{n=1}^{\infty} \frac{1}{n^2}
$$
Key Concepts
Direct Comparison TestConvergent SeriesP-SeriesInfinite Series
Direct Comparison Test
The Direct Comparison Test is a handy tool for determining whether an infinite series is likely to converge or diverge by comparing it to another series whose behavior (convergence or divergence) is already known.
Consider two series, \( \sum_{n=1}^{\infty} a_n \) and \( \sum_{n=1}^{\infty} b_n \), where each term \( a_n \) and \( b_n \) are nonnegative. If we can establish that \( a_n \leq b_n \) for all \( n \) and we know that \( \sum_{n=1}^{\infty} b_n \) converges, then we can conclude that \( \sum_{n=1}^{\infty} a_n \) also converges. Conversely, if \( a_n \geq b_n \) and \( \sum_{n=1}^{\infty} b_n \) diverges, then \( \sum_{n=1}^{\infty} a_n \) must also diverge.
Consider two series, \( \sum_{n=1}^{\infty} a_n \) and \( \sum_{n=1}^{\infty} b_n \), where each term \( a_n \) and \( b_n \) are nonnegative. If we can establish that \( a_n \leq b_n \) for all \( n \) and we know that \( \sum_{n=1}^{\infty} b_n \) converges, then we can conclude that \( \sum_{n=1}^{\infty} a_n \) also converges. Conversely, if \( a_n \geq b_n \) and \( \sum_{n=1}^{\infty} b_n \) diverges, then \( \sum_{n=1}^{\infty} a_n \) must also diverge.
Convergent Series
A Convergent Series is an infinite series that approaches a specific value as you add more terms. This means that the series has a finite sum even though there are an infinite number of terms. In simple terms, as you go further out in the series, the amount you're adding each time becomes so small that it doesn’t significantly change the total.
An important aspect of convergent series is their properties and the different tests we can use to determine convergence, including the Direct Comparison Test, Ratio Test, and Root Test, among others. Being able to identify a convergent series is crucial in mathematics, especially when dealing with functions that can be represented as power series.
An important aspect of convergent series is their properties and the different tests we can use to determine convergence, including the Direct Comparison Test, Ratio Test, and Root Test, among others. Being able to identify a convergent series is crucial in mathematics, especially when dealing with functions that can be represented as power series.
P-Series
A P-Series is a type of series with a general form of \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant. The test to determine the convergence of a p-series is straightforward:
If \( p > 1 \), the p-series is convergent. Conversely, if \( p \leq 1 \), the p-series is divergent. This test stems from the integral test, which shows that as \( p \) increases, the terms of the series become smaller faster, and the sum of the series therefore tends toward a finite limit.
If \( p > 1 \), the p-series is convergent. Conversely, if \( p \leq 1 \), the p-series is divergent. This test stems from the integral test, which shows that as \( p \) increases, the terms of the series become smaller faster, and the sum of the series therefore tends toward a finite limit.
Infinite Series
An Infinite Series is the sum of the terms of an infinite sequence. It’s written in the form \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the nth term in the sequence. The concept of infinity here does not mean that the sum is necessarily infinite; rather, the number of terms being added is infinite.
There are various types of infinite series, including convergent and divergent series, as well as specific forms such as geometric series, harmonic series, and p-series. Understanding the convergence or divergence of an infinite series is fundamental in fields such as calculus, analysis, and applied mathematics as it applies to functions and calculations involving an infinite number of terms.
There are various types of infinite series, including convergent and divergent series, as well as specific forms such as geometric series, harmonic series, and p-series. Understanding the convergence or divergence of an infinite series is fundamental in fields such as calculus, analysis, and applied mathematics as it applies to functions and calculations involving an infinite number of terms.
Other exercises in this chapter
Problem 27
(a) compute as many terms of the sequence of partial sums, \(S_{n}\), as is necessary to convince yourself that the series converges or diverges. If it converge
View solution Problem 27
Find the radius of convergence and the interval of convergence of the power series. $$ \sum_{n=1}^{\infty} \frac{2 \cdot 4 \cdot 6 \cdots \cdot 2 n}{3 \cdot 5 \
View solution Problem 28
Use the power series representations of functions established in this section to find the Taylor series of \(f\) at the given value of \(c .\) Then find the rad
View solution Problem 28
Determine whether the series is convergent, absolutely convergent, conditionally convergent, or divergent. \(\sum_{n=1}^{\infty}(-1)^{n} n \sin \left(\frac{\pi}
View solution