Chapter 1: Problem 5
Test for convergence: \(\sum_{n=2}^{\infty} \frac{(n-1)^{2}}{1+n^{2}}\)
Short Answer
Expert verified
The series diverges.
Step by step solution
01
Identify the series
Given the series to test for convergence: \( \sum_{n=2}^{\infty} \frac{(n-1)^{2}}{1+n^{2}} \).
02
Apply the Limit Comparison Test
Choose a comparison series. For large values of n, \( (n-1)^{2} \approx n^2 \) and \( 1+n^2 \approx n^2 \). Thus, we compare it with the series \( \sum_{n=2}^{\infty} \frac{n^2}{n^2} = \sum_{n=2}^{\infty} 1 \).
03
Analyze the comparison series
The comparison series \( \sum_{n=2}^{\infty} 1 \) is a divergent series, as it grows without bound (it is a p-series with p = 0 which is less than 1).
04
Calculate the limit for the Limit Comparison Test
Evaluate \( L = \lim_{{n \to \infty}} \frac{a_n}{b_n} \) where \( a_n = \frac{(n-1)^2}{1+n^2} \) and \( b_n = \frac{1}{1} = 1 \). Thus, \( L = \lim_{{n \to \infty}} \frac{(n-1)^2}{1+n^2} \cdot 1 = \lim_{{n \to \infty}} \frac{(n-1)^2}{1+n^2} \).
05
Simplify the limit expression
Factor out \( n^2 \) in both the numerator and the denominator: \( L = \lim_{{n \to \infty}} \frac{(n-1)^2}{1+n^2} = \lim_{{n \to \infty}} \frac{n^2 - 2n + 1}{n^2 + 1} = \lim_{{n \to \infty}} \frac{1 - \frac{2}{n} + \frac{1}{n^2}}{1 + \frac{1}{n^2}} \).
06
Evaluate the limit
As \( n \to \infty \), the terms involving \( \frac{1}{n} \) and \( \frac{1}{n^2} \) approach 0. Thus, \( L = \lim_{{n \to \infty}} \frac{1 - \frac{2}{n} + \frac{1}{n^2}}{1 + \frac{1}{n^2}} = \frac{1 - 0 + 0}{1 + 0} = 1 \).
07
Conclude with the Limit Comparison Test
Since \( L = 1 \) and the series \( \sum_{n=2}^{\infty} 1 \) diverges, by the Limit Comparison Test, the given series \( \sum_{n=2}^{\infty} \frac{(n-1)^{2}}{1+n^{2}} \) also diverges.
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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 useful tool for determining the convergence or divergence of an infinite series. It's especially handy when you can compare your given series to a well-known benchmark series. Here's how it works: you choose a comparison series that behaves similarly to your given series for large values of n. Then, you calculate the limit of the ratio of their corresponding terms as n tends towards infinity. Formally, if you have two series, \( \textstyle \sum a_n \) and \( \textstyle \sum b_n \), you compute \( \textstyle L = \lim_{{n \to \infty}} \frac{a_n}{b_n} \). If \( \textstyle L \) is a finite, positive number, say L > 0, then both series either converge or diverge together. This test simplifies complex series comparisons by reducing them to a manageable limit evaluation. By focusing on dominant terms that influence the behavior of the series as n becomes very large, this method offers a reliable way to draw conclusions about the given series.
Divergent Series
A series is said to diverge if its terms do not approach a finite limit or if the sum of its terms grows without bound as you take more and more terms. In simpler terms, if the series doesn’t settle down to a specific number, it diverges. The series \( \textstyle \sum_{n=2}^{\infty} 1 \) is a classic example of divergence. No matter how many terms you add, the sum keeps increasing indefinitely. In mathematical terms, this series is sometimes referred to as a “harmonic series for p=0,” where p-series with p ≤ 1 are known to diverge. Knowing whether a series diverges is crucial because it can give you insight into the behavior of more complex series when conducting comparative analysis, like with the Limit Comparison Test. Recognizing a divergent series allows you to make informed conclusions about a series you are testing for convergence or divergence.
p-Series
A p-series is a series of the form \( \textstyle \sum_{n=1}^{\infty} \frac{1}{n^p} \), where p is a positive real number. The convergence or divergence of a p-series depends entirely on the value of p:
- If p > 1, the series converges.
- If p ≤ 1, the series diverges.