Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Determine convergence or divergence for each of the series. Indicate the test you use. \(\frac{1}{2^{2}}+\frac{2}{3^{2}}+\frac{3}{4^{2}}+\frac{4}{5^{2}}+\cdots\)

Step-by-Step Solution

Answer

The series diverges by the Limit Comparison Test.

Step 1: Identify the Form of the Series

The series given is \( \sum_{n=1}^{\infty} \frac{n}{(n+1)^2} \). It resembles a standard form where each term is of the form \( \frac{n}{n^2} \) when simplified.

Step 2: Apply the Limit Comparison Test

To determine convergence, we'll use the Limit Comparison Test. Compare the series \( \sum_{n=1}^{\infty} \frac{n}{(n+1)^2} \) with the p-series \( \sum_{n=1}^{\infty} \frac{1}{n} \), a known divergent series.

Step 3: Find the Limit of the Ratio of Terms

Calculate the limit \( \lim_{n \to \infty} \frac{\frac{n}{(n+1)^2}}{\frac{1}{n}} \). Simplifying the expression gives \( \lim_{n \to \infty} \frac{n^2}{(n+1)^2} \).

Step 4: Evaluate the Limit

Simplify \( \frac{n^2}{(n+1)^2} \) to \( \frac{1}{(1+\frac{1}{n})^2} \). As \( n \to \infty \), \( \frac{1}{n} \to 0 \), and the expression approaches 1. Thus, the limit is 1.

Step 5: Conclude based on the Limit Comparison Test

Since the limit is a finite positive number (1), both series either converge or diverge together. Since the comparison series \( \sum_{n=1}^{\infty} \frac{1}{n} \) diverges, our original series \( \sum_{n=1}^{\infty} \frac{n}{(n+1)^2} \) also diverges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a handy tool used to determine whether a series converges or diverges. It works by comparing a given series to another series with known behavior. For this test, the series you want to analyze is called the test series, and you compare it to a reference series, which should ideally have similar characteristics.

Here's what you do: compute the limit of the ratio of the terms of your test series and the reference series. That is, if you have the series \( \sum a_n \) and you compare it with \( \sum b_n \), you need to find
\[ \lim_{n o \infty} \frac{a_n}{b_n}. \]
This limit, denoted as \( \lim_{n o \infty} \frac{a_n}{b_n} = L \), plays a crucial role:

• If \( L \) is a positive, finite number (like 1), then both series \( \sum a_n \) and \( \sum b_n \) either converge or diverge together.
• If \( L = 0 \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• If \( L = \infty \) and \( \sum b_n \) diverges, then \( \sum a_n \) diverges as well.

Thus, by selecting an appropriate reference series, you can infer the behavior of the series under consideration, simplifying what might otherwise be a complex task.

p-series

A p-series is a specific kind of series denoted as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant. Understanding p-series is essential because they serve as reference points in several convergence tests, including the Limit Comparison Test.

The convergence of a p-series depends on the value of \( p \):

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

In the Limit Comparison Test, p-series help to establish the convergence or divergence of more complex series. They provide a clear threshold: for example, \( \sum_{n=1}^{\infty} \frac{1}{n} \) (a Harmonic series) diverges because it's a p-series with \( p = 1 \).

Using p-series as a benchmark, you can analyze series that may not initially appear obvious. By manipulating the series into a form that resembles a p-series or can be compared to a p-series, you streamline the process of determining convergence.

Limit Evaluation

Limit evaluation is a mathematical process by which you simplify expressions to find the behavior of a sequence or function as its input approaches a particular value. This technique is crucial in analyzing series and testing convergence using the Limit Comparison Test.

For example, consider the expression \( \lim_{n o \infty} \frac{n^2}{(n+1)^2} \). To evaluate this limit, simplify each component by factoring or dividing by the highest power of \( n \) in the denominator:
\[ \lim_{n o \infty} \frac{n^2}{(n+1)^2} = \lim_{n o \infty} \frac{1}{(1+\frac{1}{n})^2}. \]
As \( n \) approaches infinity, \( 1/n \) approaches zero, making the expression
\[ \frac{1}{(1+0)^2} = 1. \]
This limit tells you that the ratio of the terms in the original series to the reference series approaches a constant value, confirming their shared behavior.

Mastering limit evaluation means you can tackle convergence tests with greater confidence, ensuring a deep understanding of the underlying patterns in series.