Problem 1 - Thomas Calculus

Question

In Exercises \(1-8,\) use the Direct Comparison Test to determine if each series converges or diverges. $$\sum_{n=1}^{\infty} \frac{1}{n^{2}+30}$$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{1}{n^{2}+30} \) converges by the Direct Comparison Test.

Step 1: Identify the Series and Comparison

We have the series \( \sum_{n=1}^{\infty} \frac{1}{n^{2}+30} \). To use the Direct Comparison Test, we need to compare this series with a known series. Here, we choose \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), a convergent \(p\)-series with \(p=2\).

Step 2: Set Up the Inequality for Comparison

For \(n \geq 1\), we observe that \(n^2 + 30 > n^2\). Therefore, \(\frac{1}{n^2 + 30} < \frac{1}{n^2}\). We will use this inequality for the Direct Comparison Test.

Step 3: Apply the Direct Comparison Test

Since \( \frac{1}{n^2 + 30} < \frac{1}{n^2} \) and the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges (as a \(p\)-series with \(p = 2 > 1\)), by the Direct Comparison Test, the series \( \sum_{n=1}^{\infty} \frac{1}{n^2 + 30} \) also converges.

Key Concepts

Convergence

In mathematics, understanding the concept of convergence is key to analyzing infinite series. Convergence refers to a series approaching a certain finite value as more and more terms are added up. When we say a series converges, it implies that as we keep summing the terms of the series, the total sum approaches a specific number and will get closer to this number, without keeping increasing indefinitely.

This is important as not all series converge. Some series, when summed, will continue to grow without bound. Such series are said to diverge. In the problem you're working on, showing convergence means proving that our series \( \sum_{n=1}^{\infty} \frac{1}{n^2+30} \) results in a finite limit when all its terms are summed indefinitely.

P-Series

A p-series is a special kind of infinite series expressed as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) where \(p\) is a positive real number. P-series are crucial because their convergence depends solely on the value of \(p\).

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

In the given exercise, the comparison series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a p-series where \(p = 2\). Since 2 is greater than 1, this series converges.

The convergence of the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) serves as a benchmark for determining the behavior of other series via the Direct Comparison Test.

Inequality

Inequalities are used to compare the terms of two different series. In our context, we look for an inequality that can help us use a known series to analyze the given series. In the Direct Comparison Test, we seek to show that the given series' term is always less than or greater than a corresponding known term from a comparison series.

Here, we used \(n^2 + 30 > n^2\) for each term in our series. This implies \( \frac{1}{n^2 + 30} < \frac{1}{n^2} \) for all values of \(n\). This strict inequality is crucial because it helps establish that the initial series is less than the known convergent series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), thereby proving convergence.

Series Comparison

Series comparison is a method used to deduce the convergence or divergence of a series whose direct evaluation might be challenging. The Direct Comparison Test is one such method. By comparing the series of interest with another series whose behavior is already known, usually a known convergent p-series or geometric series, we can infer the behavior of the original series.

• Choose a comparison series with similar terms.
• Establish a relationship using inequalities that show one series is less than or greater than the other for all terms.
• If the comparison series converges and the original series is less than it, then the original series also converges.

In your exercise, by establishing that \( \sum_{n=1}^{\infty} \frac{1}{n^2+30} \) is always less than the convergent series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), we conclusively say it converges.