Problem 17
Question
Determine convergence or divergence for each of the series. Indicate the test you use. $$ \sum_{n=1}^{\infty} \frac{4 n^{3}+3 n}{n^{5}-4 n^{2}+1} $$
Step-by-Step Solution
Verified Answer
The series converges by the Limit Comparison Test with \( \sum \frac{1}{n^2} \).
1Step 1: Identify the Series Type
We have the series \( \sum_{n=1}^{\infty} \frac{4n^3 + 3n}{n^5 - 4n^2 + 1} \). This is an infinite series with rational expressions in terms of \( n \).
2Step 2: Compare Leading Terms for Dominance
To analyze the behavior of the series, we'll compare the degrees of the polynomial in the numerator and the denominator. The leading term in the numerator is \( 4n^3 \) and in the denominator, it's \( n^5 \).
3Step 3: Asymptotic Analysis
For large \( n \), higher-degree terms dominate lower-degree ones. Thus, \( \frac{4n^3}{n^5} \approx \frac{4}{n^2} \). This suggests that for large \( n \), the series behaves similarly to \( \sum \frac{4}{n^2} \).
4Step 4: Apply the Limit Comparison Test
Compare \( a_n = \frac{4n^3 + 3n}{n^5 - 4n^2 + 1} \) with \( b_n = \frac{1}{n^2} \). Compute \( \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{4n^3 + 3n}{n^5 - 4n^2 + 1} \times n^2 = \lim_{n \to \infty} \frac{4n^5 + 3n^3}{n^5 - 4n^2 + 1} = 4 \).
5Step 5: Determine Convergence by the Result of the Test
Since the limit in the Limit Comparison Test is a positive finite number (4), and \( \sum \frac{1}{n^2} \) is a convergent p-series (\(p = 2\)), \( \sum \frac{4n^3 + 3n}{n^5 - 4n^2 + 1} \) also converges.
Key Concepts
Infinite SeriesLimit Comparison Testp-Series Convergence
Infinite Series
An infinite series is the sum of an infinite sequence of numbers. Essentially, it involves adding terms endlessly, which can either approach a certain finite value or continue to grow without bound. The most common representation is using the summation notation, such as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the terms in the series.
A series is said to be convergent if the sequence of its partial sums approaches a fixed number as more terms are added. If the sums do not stabilize to any number, the series is divergent.
A series is said to be convergent if the sequence of its partial sums approaches a fixed number as more terms are added. If the sums do not stabilize to any number, the series is divergent.
- A convergent series leads to a specific value, meaning the infinite "sum" settles on a finite number.
- A divergent series either increases indefinitely or doesn't settle on any specific number.
Limit Comparison Test
The Limit Comparison Test is a handy tool for analyzing the convergence of infinite series. It works by comparing a given series to another series whose convergence behavior is known. The idea is simple: if you have a complicated series, see if it behaves similarly to a simpler series.
Here's how the Limit Comparison Test is applied:
Here's how the Limit Comparison Test is applied:
- Identify the target series \( \sum a_n \).
- Choose a simpler comparison series \( \sum b_n \) that resembles \( \sum a_n \) in behavior.
- Calculate the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} \).
p-Series Convergence
A p-series is a specific type of series where each term is of the form \( \frac{1}{n^p} \). The series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) has distinct convergence behaviors according to the value of \( p \):
The simplicity of the p-series makes them easy to analyze and apply in various convergence tests. It's widely used in mathematical analysis, calculus, and even in fields like physics and engineering where understanding infinite processes is crucial. By linking the given series to the behavior of a p-series, we can simplify and resolve complex series to assess their convergence swiftly.
- If \( p > 1 \), the p-series converges.
- If \( p \leq 1 \), the p-series diverges.
The simplicity of the p-series makes them easy to analyze and apply in various convergence tests. It's widely used in mathematical analysis, calculus, and even in fields like physics and engineering where understanding infinite processes is crucial. By linking the given series to the behavior of a p-series, we can simplify and resolve complex series to assess their convergence swiftly.
Other exercises in this chapter
Problem 17
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Classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=2}^{\infty}(-1)^{n} \frac{1}{n \ln n} $$
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Write the given decimal as an infinite series, then find the sum of the series, and finally, use the result to write the decimal as a ratio of two integers (see
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Use any test developed so far, including any from Section \(9.2\), to decide about the convergence or divergence of the series. Give a reason for your conclusio
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