Chapter 9: Problem 20
Indicate whether the given series converges or diverges and give a reason for your conclusion. $$\sum_{n=1}^{\infty} \frac{n+5}{1+n^{3}}$$
Short Answer
Expert verified
The series converges by the limit comparison test with \( \sum \frac{1}{n^2} \).
Step by step solution
01
Analyze the General Term
The general term of the series is given by \( a_n = \frac{n+5}{1+n^3} \). As \( n \to \infty \), \( a_n \to 0 \) because the highest power of \( n \) in the denominator (\( n^3 \)) grows faster than any term in the numerator (\( n \)).
02
Apply the Limit Comparison Test
Compare the series \( \sum \frac{n+5}{1+n^3} \) with the simpler series \( \sum \frac{1}{n^2} \), which is a p-series with \( p = 2 > 1 \) and therefore converges. Consider the ratio \( \lim_{n \to \infty} \frac{\frac{n+5}{1+n^3}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2(n+5)}{1+n^3} \).
03
Simplify the Limit
Simplify \( \lim_{n \to \infty} \frac{n^2(n+5)}{1+n^3} = \lim_{n \to \infty} \frac{n^3 + 5n^2}{1 + n^3} \). Dividing the numerator and the denominator by \( n^3 \), we get \( \lim_{n \to \infty} \frac{1 + \frac{5}{n}}{\frac{1}{n^3} + 1} \). As \( n \to \infty \), this simplifies to \( \frac{1}{1} = 1 \).
04
Conclude Using the Limit Comparison Test
Since the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} = 1 \) and \( \sum \frac{1}{n^2} \) converges, by the limit comparison test, \( \sum \frac{n+5}{1+n^3} \) also converges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Comparison Test
The Limit Comparison Test is a valuable tool for determining the convergence or divergence of series. It works by comparing a given series to a well-known benchmark series. The beauty of this test lies in its simplicity and effectiveness when dealing with complex series.
Here’s how it works: Consider two series, \( \sum a_n \) and \( \sum b_n \). You choose \( b_n \) such that it resembles \( a_n \) in behavior as \( n \to \infty \). This often involves selecting a \( p\)-series since their convergence properties are well-understood. You then calculate the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \).
The Limit Comparison Test simplifies your task by letting you use a simpler series to understand a more complicated one. This comparison is often directly to a \( p\)-series, making it a frequently used technique in calculus.
Here’s how it works: Consider two series, \( \sum a_n \) and \( \sum b_n \). You choose \( b_n \) such that it resembles \( a_n \) in behavior as \( n \to \infty \). This often involves selecting a \( p\)-series since their convergence properties are well-understood. You then calculate the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \).
- If this limit is a positive finite number, then \( a_n \) and \( b_n \) either both converge or both diverge.
- If the limit is zero and \( \sum b_n \) converges, then \( \sum a_n \) converges.
- If the limit is infinite and \( \sum b_n \) diverges, then \( \sum a_n \) diverges.
The Limit Comparison Test simplifies your task by letting you use a simpler series to understand a more complicated one. This comparison is often directly to a \( p\)-series, making it a frequently used technique in calculus.
P-series
A \( p\)-series is a specific type of series in the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a constant. Understanding \( p\)-series is crucial because they serve as a fundamental building block for analyzing other series.
The convergence of a \( p\)-series depends solely on the value of \( p \):
For instance, the series \( \sum \frac{1}{n^2} \) is a \( p\)-series with \( p = 2 \). Since \( 2 > 1 \), this series converges. It's pivotal as a benchmark in convergence tests. In practice, many complex series can be reduced to a \( p\)-series through simplification, making it easier to ascertain their convergence properties.
The convergence of a \( p\)-series depends solely on the value of \( p \):
- If \( p > 1 \), the series converges. This means that the infinite sum approaches a finite value.
- If \( 0 < p \leq 1 \), the series diverges. Therefore, the terms add up to an infinite value as \( n \to \infty \).
For instance, the series \( \sum \frac{1}{n^2} \) is a \( p\)-series with \( p = 2 \). Since \( 2 > 1 \), this series converges. It's pivotal as a benchmark in convergence tests. In practice, many complex series can be reduced to a \( p\)-series through simplification, making it easier to ascertain their convergence properties.
Convergence and Divergence of Series
In the world of infinite series, determining whether a series converges or diverges is a core challenge. The terms **convergence** and **divergence** indicate whether the sums of the terms of a series approach a finite limit or not.
Understanding whether a series converges or diverges provides insights into the behavior of the sequence of terms being summed. It's a foundational concept that underpins much of mathematical analysis. Using tools like the Limit Comparison Test, you can make this determination by comparing your series to others whose behavior is well-known.
- **Convergence:** A series \( \sum a_n \) converges if the sequence of partial sums \( S_n = a_1 + a_2 + ... + a_n \) approaches a limiting value as \( n \to \infty \). This means the series adds up to a specific, finite number.
- **Divergence:** Conversely, a series diverges if the sequence of partial sums does not approach a finite limit. The series keeps growing without bound or oscillates.
Understanding whether a series converges or diverges provides insights into the behavior of the sequence of terms being summed. It's a foundational concept that underpins much of mathematical analysis. Using tools like the Limit Comparison Test, you can make this determination by comparing your series to others whose behavior is well-known.