Problem 20 - Calculus: Early Transcendentals

Question

Determine whether the series is convergent or divergent. \( \displaystyle \sum_{n = 3}^{\infty} \frac {3n - 4}{n^2 - 2n} \)

Step-by-Step Solution

Answer

The series \( \sum_{n=3}^{\infty} \frac{3n - 4}{n^2 - 2n} \) is divergent.

Step 1: Identify the series and simplify the expression

The given series is \( \sum_{n=3}^{\infty} \frac{3n - 4}{n^2 - 2n} \). First, simplify the series term \( \frac{3n - 4}{n^2 - 2n} \). Factor the denominator as \( n(n-2) \), which gives the simplified term as \( \frac{3n - 4}{n(n-2)} \).

Step 2: Perform the limit comparison test for convergence

To apply the limit comparison test, compare the given series \( \sum_{n=3}^{\infty} \frac{3n - 4}{n(n-2)} \) with a known simple series. Consider the series \( \sum_{n=3}^{\infty} \frac{1}{n} \), which is a p-series with \( p = 1 \) and is known to be divergent.

Step 3: Calculate the limit of the comparison

Determine the limit \( L = \lim_{n \to \infty} \frac{\frac{3n - 4}{n(n-2)}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{3n^2 - 4n}{n^2 - 2n^2} = \lim_{n \to \infty} \frac{3n^2 - 4n}{-2n^2} \). Simplify this to \( L = \lim_{n \to \infty} \frac{3 - \frac{4}{n}}{-2} = -\frac{3}{2} \).

Step 4: Analyze the limit to determine the series behavior

Since the limit \( L = -\frac{3}{2} \) is a finite non-zero number, both series \( \sum_{n=3}^{\infty} \frac{3n - 4}{n(n-2)} \) and \( \sum_{n=3}^{\infty} \frac{1}{n} \) either converge or diverge together. Since \( \sum_{n=3}^{\infty} \frac{1}{n} \) is a harmonic series (divergent), the given series is also divergent.

Key Concepts

Limit Comparison Test

The limit comparison test (LCT) is a method used to determine the convergence or divergence of an infinite series by comparing it to another series whose behavior is already known. The main idea is to compare a complicated series to a simpler one, like a p-series or a harmonic series, and use the limit of their ratio as the basis for the comparison.

Here's how it works:

• Given a series \( \sum a_n \,,\) we select a series \( \sum b_n \,.\)
• Calculate the limit \( L = \lim_{{n \to \infty}} \frac{{a_n}}{{b_n}} \,.\)
• If \( L \,\) is a finite, positive number then both series \( \sum a_n \,\) and \( \sum b_n \,\) have the same convergence behavior, meaning either both converge or both diverge.

It's crucial to select an appropriate \( b_n \,\) such that its convergence characteristics are already known, like a p-series with a particular value of \( p \,\) or the harmonic series, which is a common divergent series.

Simplification of Expressions

Simplifying expressions is often a preliminary step in analyzing series for convergence. By simplifying complicated fractions or algebraic terms, we can make subsequent calculations easier and clearer.

Let's consider the series term from the exercise: \( \frac{3n - 4}{n^2 - 2n} \,.\)

• Factor the denominator, \( n^2 - 2n = n(n - 2) \,\) to obtain \( \frac{3n - 4}{n(n - 2)} \,.\)
• This step helps to reveal the dominant terms as \( n \,\) becomes large, aiding in comparison with simpler series.
• If necessary, break down complex expressions further to identify leading terms which influence the series behavior as \( n o \infty\)

By simplifying, we identify what mainly drives the series' divergence or convergence, making it easier to apply tests like the limit comparison test.

P-Series

P-series are a family of series useful in convergence testing. A p-series takes the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \,.\)

• The convergence or divergence of a p-series depends directly on the value of \( p \,.\)
• If \( p > 1 \,\), the series converges.
• If \( p \leq 1 \,\), the series diverges.

P-series provide a standard against which we can compare other series, particularly using tests like the limit comparison test. They are straightforward to analyze, making them valuable in evaluating more complex series. Recognizing a series as a p-series or approximating it to one is a common technique in determining the series’ overall behavior.

Harmonic Series

The harmonic series is a specific example of a divergent p-series where \( p = 1 \,\), represented by \( \sum_{n=1}^{\infty} \frac{1}{n} \,.\)

• Despite the decreasing size of terms with increasing \( n \,\), the harmonic series diverges.
• This divergence is a classic result and a foundational example in series comparison.
• Harmonic series often serve as a comparison benchmark because of their well-known behavior.

In the given exercise, the harmonic series is used in the limit comparison test. Despite its slow growth, its divergence means any similar large-sum behavior series will likely diverge too. Understanding why the harmonic series diverges helps reveal the fate of other series when comparing, even those that superficially look different.