Question

Use the comparison test to determine whether the following series converge. $$ \sum_{n=1}^{\infty} a_{n} \text { where } a_{n}=\frac{2}{n(n+1)} $$

Short answer

The series converges by the comparison test.

Step-by-step solution

Identify the Series

We are given the series \( \sum_{n=1}^{\infty} a_{n} \) where \( a_{n} = \frac{2}{n(n+1)} \). We need to determine whether this series converges using the comparison test.

Choose a Comparison Series

The given term \( a_{n} = \frac{2}{n(n+1)} \) can be decomposed to \( \frac{2}{n^2 + n} \). For large values of \( n \), \( n(n+1) \approx n^2 \). So, let's compare this series with \( b_n = \frac{2}{n^2} \), which is a simpler form and converges because it is a p-series with \( p = 2 > 1 \).

Apply Limit Comparison Test

Using the limit comparison test, calculate \( \lim_{n \to \infty} \frac{a_n}{b_n} \). Thus, we have \[ \lim_{n \to \infty} \frac{\frac{2}{n(n+1)}}{\frac{2}{n^2}} = \lim_{n \to \infty} \frac{2n^2}{2n^2 + 2n} = \lim_{n \to \infty} \frac{n}{n+1} = 1. \] Since this limit is a positive finite number (in this case, 1), the behavior of the series \( \sum a_n \) is the same as the series \( \sum b_n \).

Conclude Convergence

Since \( \sum b_n = \sum \frac{2}{n^2} \) converges (it is a p-series with \( p = 2 > 1 \)), by the limit comparison test, the original series \( \sum a_n = \sum \frac{2}{n(n+1)} \) also converges.

Key concepts

Convergence of Series

Understanding the convergence of series is a fundamental part of analyzing infinite series in calculus. A series consists of adding up terms in a sequence, and it converges if the sum approaches a finite number as more terms are added. Convergence is different from divergence, where the sum keeps growing indefinitely without approaching a particular value.

To determine convergence, various tests can be applied, and each test uses different criteria based on the properties of the series. The Comparison Test is one such method. It is commonly used because it compares a given series to a known benchmark series, typically simpler to analyze. When a complicated series behaves similarly to a known convergent series, we conclude that it converges as well.

Hence, convergence implies that we can assign a specific value to the sum of an infinite series, crucial for applications in mathematics and applied sciences.

P-Series

A P-Series is a special type of series that takes the form: \[ \sum_{n=1}^{\infty} \frac{1}{n^p} \]The behavior of a P-Series heavily depends on the value of the exponent \( p \). When analyzing whether a P-Series converges, the critical factor is whether \( p \) is greater than, equal to, or less than one.

• If \( p > 1 \), the P-Series converges.
• If \( p \leq 1 \), the P-Series diverges.

This makes P-Series a useful reference in the Comparison Test, since their convergence is well-defined and easy to recognize. In our original exercise, the series \( \sum \frac{2}{n^2} \) has \( p = 2 \). It converges because \( p > 1 \). This is a classic convergent P-Series, making it an ideal choice for comparison with similar, but more complex, series.

Limit Comparison Test

The Limit Comparison Test is an important tool used in determining whether a series converges or diverges. It is particularly useful when direct comparison is difficult, as it allows for using limits.

The test involves comparing two series: \( \sum a_n \) and \( \sum b_n \). To apply it, calculate the limit:\[\lim_{n \to \infty} \frac{a_n}{b_n}\]Here’s how it works:

• If the limit is a positive finite number, then both series either converge or diverge together.
• If the limit is zero or infinity, the test is inconclusive.

In our case, the original series \( \sum \frac{2}{n(n+1)} \) was compared to a simpler P-Series \( \sum \frac{2}{n^2} \). By calculating the limit of their ratio as \( n \) goes to infinity, we found it equals 1—a positive finite number. This confirmed that both series share the same behavior: since the P-Series converges, so does the initial series.