Question

Test for convergence: \(\sum_{n=2}^{\infty} \frac{(n-1)^{2}}{1+n^{2}}\)

Short answer

The series diverges.

Step-by-step solution

Identify the series

Given the series to test for convergence: \( \sum_{n=2}^{\infty} \frac{(n-1)^{2}}{1+n^{2}} \).

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 \( \sum_{n=2}^{\infty} \frac{n^2}{n^2} = \sum_{n=2}^{\infty} 1 \).

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).

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, \( L = \lim_{{n \to \infty}} \frac{(n-1)^2}{1+n^2} \cdot 1 = \lim_{{n \to \infty}} \frac{(n-1)^2}{1+n^2} \).

Simplify the limit expression

Factor out \( n^2 \) in both the numerator and the denominator: \( L = \lim_{{n \to \infty}} \frac{(n-1)^2}{1+n^2} = \lim_{{n \to \infty}} \frac{n^2 - 2n + 1}{n^2 + 1} = \lim_{{n \to \infty}} \frac{1 - \frac{2}{n} + \frac{1}{n^2}}{1 + \frac{1}{n^2}} \).

Evaluate the limit

As \( n \to \infty \), the terms involving \( \frac{1}{n} \) and \( \frac{1}{n^2} \) approach 0. Thus, \( L = \lim_{{n \to \infty}} \frac{1 - \frac{2}{n} + \frac{1}{n^2}}{1 + \frac{1}{n^2}} = \frac{1 - 0 + 0}{1 + 0} = 1 \).

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.

Key concepts

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.

For instance, the series \( \textstyle \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges because p = 2, which is greater than 1. Conversely, the series \( \textstyle \sum_{n=1}^{\infty} \frac{1}{n} \) diverges as p = 1. The simplicity of p-series makes them great candidates for comparison tests. By comparing the behavior of a more complex series to a simple p-series, one can often determine the convergence or divergence in a more straightforward manner. p-Series offer a robust framework for tackling a wide variety of convergence problems, particularly when leveraged in tests like the Limit Comparison Test.