Problem 52
Question
Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{\sqrt[n]{n}}{n^{2}}\end{equation}
Step-by-Step Solution
Verified Answer
The series \( \sum_{n=1}^{\infty} \frac{\sqrt[n]{n}}{n^{2}} \) converges.
1Step 1: Identify the series
We are given the series \( \sum_{n=1}^{\infty} \frac{\sqrt[n]{n}}{n^{2}} \). Our task is to determine if this series converges or diverges.
2Step 2: Simplify the General Term
The general term of our series is \( a_n = \frac{\sqrt[n]{n}}{n^{2}} \). The expression \( \sqrt[n]{n} \) can be rewritten using exponents as \( n^{1/n} \), leading to the term becoming \( \frac{n^{1/n}}{n^2} \).
3Step 3: Estimate Limit of Simplified Term
As \( n \) approaches infinity, the term \( n^{1/n} \) approaches 1 because any number raised to power \(1/n \) approaches 1. Therefore, \( \lim_{n \to \infty} n^{1/n} = 1 \).
4Step 4: Analyze the Behavior of the Term
Substituting the limit found in the previous step, the term \( a_n \) becomes \( \frac{1}{n^2} \) for large \( n \).
5Step 5: Use the p-Series Test
The series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a p-series with \( p = 2 > 1 \). According to the p-series convergence test, a series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges if \( p > 1 \).
6Step 6: Conclusion on Convergence
Since the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges, and the general term of our series behaves like \( \frac{1}{n^2} \) for large \( n \), the series \( \sum_{n=1}^{\infty} \frac{\sqrt[n]{n}}{n^{2}} \) converges as well.
Key Concepts
Understanding the p-series testExploring the limit comparison testUnderstanding infinite series
Understanding the p-series test
The p-series test is a fundamental method used in determining the convergence or divergence of infinite series. It's especially useful when dealing with series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). Here, the key variable is \( p \), known as the power of the series.
- If \( p > 1 \), the series converges. This means that as we sum up an infinite number of terms, the total approaches a finite number.
- If \( p \leq 1 \), the series diverges. This means as we continue to add terms from 1 to infinity, the total sum goes to infinity.
Exploring the limit comparison test
The limit comparison test is another crucial method for analyzing series. It helps determine the convergence of a particular series by comparing it to another series whose convergence behavior is already known. Here's how the test works:
- Start with two series, \( \sum a_n \) and \( \sum b_n \), whose terms are positive.
- Calculate the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} \).
- If \( 0 < L < \infty \), then either both series converge, or both diverge. This means both series have the same behavior concerning convergence.
Understanding infinite series
An infinite series is a sum of an infinite sequence of numbers. It's an essential concept in mathematics with numerous applications in calculus, physics, and engineering. Understanding the basics can simplify approaches to complex mathematical challenges.An infinite series is written as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the individual terms.
Types of Series
- Convergent Series: These have a sum that approaches a finite number as more terms are added. The prime example is the p-series with \( p > 1 \).
- Divergent Series: These series sum to infinity or do not settle to a specific value. For instance, the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \).
Other exercises in this chapter
Problem 52
How many terms of the convergent series \(\sum_{n=4}^{\infty}\left(1 / n(\ln n)^{3}\right)\) should be used to estimate its value with error at most 0.01\(?\)
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