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

The series given is \[\begin{equation} onumber \ \ \sum_{n=2}^{\infty} \frac{(n-1)^{2}}{1+n^{2}} \end{equation}\]

Choose a Convergence Test

To determine the convergence of the series, use the Limit Comparison Test, which is useful for rational functions.

Find a Comparable Series

Compare \( \frac{(n-1)^{2}}{1+n^{2}} \) with \( \frac{n^{2}}{n^{2}} = 1 \). For large values of n, \( \frac{(n-1)^{2}}{1+n^{2}} \approx \frac{n^{2}}{n^{2}} \approx 1 \).

Apply the Limit Comparison Test

Calculate the limit \( \lim_{{n \to \infty}} \frac{\frac{(n-1)^{2}}{1+n^{2}}}{1} = \lim_{{n \to \infty}} \frac{(n-1)^{2}}{1+n^{2}} \).Simplify the expression: \( \lim_{{n \to \infty}} \frac{(n-1)^{2}}{1+n^{2}} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n^{2}+1} \).

Evaluate the Limit

By dividing the numerator and denominator by \( n^{2} \), we get:\( \lim_{{n \to \infty}} \frac{1-\frac{2}{n} + \frac{1}{n^{2}}}{1+\frac{1}{n^{2}}} \).As \( n \) approaches infinity, the terms \( \frac{2}{n} \ and \ \frac{1}{n^{2}} \) approach 0. Thus, the limit is \( \lim_{{n \to \infty}} \frac{1-0+0}{1+0} = 1 \).

Determine Convergence

Since the limit comparison test yields a finite non-zero limit (1) and the series \( \sum_{n=2}^{\infty} 1 \) (a divergent harmonic series) diverges, 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 helps to determine if a series converges or diverges by comparing it to another series whose behavior is known.

This test is especially useful for rational functions. Here's how it works:

• First, identify the given series, \(\frac{(n-1)^2}{1+n^2}\).
• Select a second series, \(\frac{n^2}{n^2} = 1\), which is known to converge or diverge.
• Next, compute the limit of the ratio of the series' terms as n approaches infinity:

\[\lim_{{n \to \infty}} \frac{a_n}{b_n} = L\]Here, \(a_n\) is the term of the given series, and \(b_n\) is the term of the comparison series.

If \(L\) is a finite, non-zero constant:\

• If \(\sum b_n\) converges, then \(\sum a_n\) converges.
• If \(\sum b_n\) diverges, then \(\sum a_n\) diverges.

This test simplifies the convergence question by leveraging known series behavior.

In the exercise, the given series is compared to a divergent harmonic series \(\sum_{n=2}^{\infty} 1\).

The limit calculated is 1, indicating that the series \(\sum_{n=2}^{\infty} \frac{(n-1)^2}{1+n^2}\) diverges as well.

Rational Functions

Rational functions are ratios of polynomials. In this context, they appear in the form:

\[\frac{P(n)}{Q(n)}\]

Here, \(P(n)\) and \(Q(n)\) are polynomials.

Understanding rational functions is crucial for applying the Limit Comparison Test.

When considering large values of \(n\), we often focus on the leading terms of the polynomials. These leading terms dominate the behavior of the function. For example, in the series \(\frac{(n-1)^2}{1+n^2}\), the numerator and denominator both have leading terms involving \(n^2\).

This helps simplify the comparison series because:

• \((n-1)^2\) expands to \(n^2 - 2n + 1\), and
• The denominator is \(1 + n^2\).

For large \(n\), the terms \(2n\) and 1 become insignificant, resulting in a simplified form roughly equal to \(\frac{n^2}{n^2}\), which is 1.

Understanding how to approximate rational functions for large \(n\) is key to applying comparison tests effectively.

Divergent Series

A divergent series does not approach a finite limit as more terms are added.

An example is the harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\), which diverges.

In the exercise, the series \(\sum_{n=2}^{\infty} 1\) is used for comparison. Since each term is constant, adding infinitely many 1's results in divergence.

This behavior is tested using the Limit Comparison Test.

When the test shows that the limit of the ratio of the given series and the comparison series is a finite, non-zero constant, the given series follows the same convergence or divergence behavior as the comparison series.

In this case, since \(\sum_{n=2}^{\infty} \frac{(n-1)^2}{1+n^2}\) has a limit ratio equal to 1 against the divergent series \(\sum_{n>=2}^{\infty} 1\), it too diverges.

Recognizing divergent series helps in selecting appropriate comparison series for convergence tests and understanding the behavior of more complex series.