Problem 31
Question
In Exercises \(31-46,\) find the sum of the convergent series. $$ \sum_{n=2}^{\infty} \frac{1}{n^{2}-1} $$
Step-by-Step Solution
Verified Answer
The sum of the series is \( \frac{1}{3} \)
1Step 1: Breaking the expression into partial fractions
We start by expressing the rational function \( \frac{1}{n^{2}-1} \) as a partial fraction. We have \( n^{2}-1 = (n-1)(n+1) \). Therefore, we write: \( \frac{1}{n^{2}-1} = \frac{A}{n-1} + \frac{B}{n+1} \). Solving for A and B gives A = 1/2, B = -1/2.
2Step 2: Expression
Our infinite series now becomes \( \sum_{n=2}^{\infty} \left( \frac{1}{2(n-1)} - \frac{1}{2(n+1)}\right) \)
3Step 3: Apply telescoping series
Notice that most of the terms after the first will cancel out because of the form of the series. This is a property of a telescoping series. The series reduces to \( \frac{1}{2} - \frac{1}{6} + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{ 4}\right) + \cdots \), with all subsequent terms cancelling out.
4Step 4: Simplification
The only terms that remain after all the cancellations are the first term from the first element and the negative part of the second term of every element. i.e.: \( \frac{1}{2} - \frac{1}{6} = \frac{1}{3} \)
5Step 5: Conclusion
The sum of the series is \( \frac{1}{3} \).
Key Concepts
Partial Fraction DecompositionTelescoping SeriesInfinite SeriesSum of Series
Partial Fraction Decomposition
Partial fraction decomposition is a powerful mathematical technique used to simplify rational expressions. When you have a rational function that is difficult to work with directly, breaking it down into a sum of simpler fractions can make it easier to analyze and compute.
In the given exercise, we needed to decompose the expression \( \frac{1}{n^2-1} \). The denominator \( n^2-1 \) can be factored into \( (n-1)(n+1) \). With partial fraction decomposition, we express it as:
In the given exercise, we needed to decompose the expression \( \frac{1}{n^2-1} \). The denominator \( n^2-1 \) can be factored into \( (n-1)(n+1) \). With partial fraction decomposition, we express it as:
- \( \frac{1}{n^2-1} = \frac{A}{n-1} + \frac{B}{n+1} \)
Telescoping Series
A telescoping series is a series where terms effectively cancel each other out when added up. This is an advantage because it simplifies computation, leaving a much smaller number of terms to sum.
In the decomposed series \( \sum_{n=2}^{\infty} \left( \frac{1}{2(n-1)} - \frac{1}{2(n+1)} \right) \), each subtraction leads to cancellation with a part of the subsequent expression. More explicitly, the sum forms a pattern like \( \frac{1}{2} - \frac{1}{2} \) where these parts add to zero as you move on.
This cancellation results in the effective reduction of the series complexity, allowing you to focus only on the non-cancelled terms, crucially simplifying the calculation of an infinite series sum.
In the decomposed series \( \sum_{n=2}^{\infty} \left( \frac{1}{2(n-1)} - \frac{1}{2(n+1)} \right) \), each subtraction leads to cancellation with a part of the subsequent expression. More explicitly, the sum forms a pattern like \( \frac{1}{2} - \frac{1}{2} \) where these parts add to zero as you move on.
This cancellation results in the effective reduction of the series complexity, allowing you to focus only on the non-cancelled terms, crucially simplifying the calculation of an infinite series sum.
Infinite Series
An infinite series refers to the sum of an infinite sequence of numbers. Mathematically, it's expressed in the form \( \sum_{n=2}^{\infty} a_n \), where \( a_n \) is a sequence term indexed by \( n \).
Infinite series can converge or diverge. A convergent series approaches a finite limit, as the series progresses indefinitely. The infinite series given in the exercise converges due to the nature of its terms, especially after decomposition and cancellation through telescoping. Understanding that not all infinite series converge is vital, and various tests can determine convergence. However, the exercise series is made manageable by its structural properties.
Infinite series can converge or diverge. A convergent series approaches a finite limit, as the series progresses indefinitely. The infinite series given in the exercise converges due to the nature of its terms, especially after decomposition and cancellation through telescoping. Understanding that not all infinite series converge is vital, and various tests can determine convergence. However, the exercise series is made manageable by its structural properties.
Sum of Series
Calculating the sum of an infinite series can be complex. However, with methods like partial fraction decomposition and recognizing telescoping properties, it becomes simpler.
After identifying and using telescoping patterns, you're left with fewer terms to sum, leading to a straightforward computation.
In the exercise, after decomposing and canceling out innumerable terms through telescoping, we're left to calculate the sum of remaining non-cancelled terms, which simplifies to:
After identifying and using telescoping patterns, you're left with fewer terms to sum, leading to a straightforward computation.
In the exercise, after decomposing and canceling out innumerable terms through telescoping, we're left to calculate the sum of remaining non-cancelled terms, which simplifies to:
- \( \frac{1}{2} - \frac{1}{6} = \frac{1}{3} \)
Other exercises in this chapter
Problem 30
Use Theorem 7.15 to determine the number of terms required to approximate the sum of the series with an error of less than 0.001. $$ \sum_{n=1}^{\infty} \frac{(
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Use the series for \(f(x)=\arctan x\) to approximate the value, using \(R_{N} \leq 0.001\). \(\int_{0}^{1 / 2} x^{2} \arctan x d x\)
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Let \(f\) be a positive, continuous, and decreasing function for \(x \geq 1\) such that \(a_{n}=f(n)\). Prove that if the series \(\sum_{n=1}^{\infty} a_{n}\) c
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In Exercises \(31-34,\) find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.) $$
View solution