Question

Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=4}^{\infty} \frac{2 n}{n^{2}-9}$$

Short answer

The series diverges by the Limit Comparison Test.

Step-by-step solution

- Preliminary Test (Divergence Test)

First, apply the Divergence Test which states that if the limit of the sequence terms doesn't approach 0, the series diverges. Evaluate the limit of the general term as follows: \[ \text{Let } a_n = \frac{2n}{n^2 - 9} \] Calculate: \[ \text{lim}_{n \to \text{infty}} \frac{2n}{n^2 - 9} \] Divide the numerator and denominator by \( n \): \[ \text{lim}_{n \to \text{infty}} \frac{2}{n - \frac{9}{n}} \] As \( n \to \text{infty} \), \( \frac{9}{n} \to 0 \). Thus we have: \[ \text{lim}_{n \to \text{infty}} \frac{2}{n} = 0 \] Since the limit is 0, the Divergence Test is inconclusive.

- Choose a Convergence Test: Comparison Test

The given series resembles a rational function where the degree of polynomial in the denominator is greater than the degree in the numerator. The Comparison Test will be suitable. Compare the given series with a simpler series: \[ 0 < \frac{2n}{n^2 - 9} < \frac{2n}{n^2/2} = \frac{4}{n}, \forall n \text{ sufficiently large} \] Consider the p-series: \( \sum_{n=4}^{\text{infty}} \frac{1}{n} \) which diverges because \( p = 1 \leq 1 \).

- Application of the Limit Comparison Test

To rigorously apply the Comparison Test, use the Limit Comparison Test with \( b_n = \frac{1}{n} \): \[ \text{Let } a_n = \frac{2n}{n^2 - 9} \text{ and } b_n = \frac{1}{n} \] Compute the limit: \[ \text{lim}_{n \to \text{infty}} \frac{a_n}{b_n} = \text{lim}_{n \to \text{infty}} \frac{\frac{2n}{n^2 - 9}}{\frac{1}{n}} = \text{lim}_{n \to \text{infty}} \frac{2n^2}{n^2 - 9} = \] \[ \text{lim}_{n \to \text{infty}} \frac{2n^2}{n^2} = 2 \] Since the limit is a finite, nonzero positive number (\(2\)), by the Limit Comparison Test, \( \sum a_n \) and \( \sum b_n \) have the same behavior. Since \( \sum b_n = \sum \frac{1}{n} \) diverges, \( \sum_{n=4}^{\text{infty}} \frac{2n}{n^2 - 9} \) also diverges.

Key concepts

Divergence Test

When testing a series for convergence or divergence, it's wise to begin with the Divergence Test.
This test checks if the limit of the sequence's terms does not approach zero as n approaches infinity.
If the limit is not zero, the series diverges. For our series:
Let: \( a_n = \frac{2n}{n^2 - 9} \)

• Calculate the limit: \( \text{lim}_{n \to \text{infty}} a_n \)
• Divide each term by \( n \): \( \text{lim}_{n \to \text{infty}} \frac{2}{n - \frac{9}{n}} \)
• As \( n \to \text{infty} \), \( \frac{9}{n} \) approaches 0, giving us: \( \text{lim}_{n \to \text{infty}} \frac{2}{n} = 0 \)

Since the limit is 0, the Divergence Test is inconclusive here.

Comparison Test

When the Divergence Test is inconclusive, other tests like the Comparison Test can help.
This test involves comparing the series to a known benchmark series. Choose: \( \frac{4}{n} \) because:

• For large n, \( \frac{2n}{n^2 - 9} \) approximately behaves like \( \frac{4}{n} \)
• Comparison: \( 0 < \frac{2n}{n^2 - 9} < \frac{4}{n} \)

Given the benchmark p-series: \( \frac{1}{n} \) diverges because exponent p = 1.
If the series being compared diverges, the original series also diverges.

Limit Comparison Test

For a stricter comparison, use the Limit Comparison Test. This test compares two series by computing the limit of their ratio.
If the limit is a finite, non-zero constant, both series either converge or diverge together.
Consider:

• \( a_n = \frac{2n}{n^2 - 9} \)
• \( b_n = \frac{1}{n} \)

Calculate the limit: \( \text{lim}_{n \to \text{infty}} \frac{a_n}{b_n} \)

• Combine ratios: \( \text{lim}_{n \to \text{infty}} \frac{\frac{2n}{n^2 - 9}}{\frac{1}{n}} = \frac{2n^2}{n^2 - 9} = \frac{2n^2}{n^2} = 2 \)

The limit being 2 (non-zero, positive) means both series, \( \frac{2n}{n^2 - 9} \) and \( \frac{1}{n} \) have the same divergence behavior.
Since \( \frac{1}{n} \) diverges, so does \( \frac{2n}{n^2 - 9} \).

p-series

The p-series is a significant concept in understanding series convergence.
A p-series is of the form \( \frac{1}{n^p} \). Here's how convergence depends on p:

• If \( p > 1 \), the series converges.
• If \( p \le 1 \), the series diverges.

For example, the harmonic series \( \frac{1}{n} \) is a p-series where p = 1.
It diverges because p \( \le 1 \).
Recognizing and comparing against p-series helps determine the behavior of more complex series.