Problem 6 - Thomas Calculus

Question

Use the Comparison Test to determine if each series converges or diverges. \begin{equation}\sum_{n=1}^{\infty} \frac{1}{n 3^{n}}\end{equation}

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{1}{n 3^n} \) converges.

Step 1: Identify the given series

The given series is \( \sum_{n=1}^{\infty} \frac{1}{n 3^{n}} \). We need to determine whether this series converges or diverges using the Comparison Test.

Step 2: Choose a Comparison Series

We choose a geometrically simpler series for comparison. Since the denominator has \(3^n\), which resembles a geometric series, consider \( \sum_{n=1}^{\infty} \frac{1}{3^{n}} \), which is a geometric series with common ratio \( \frac{1}{3} \).

Step 3: Determine Convergence of Comparison Series

The series \( \sum_{n=1}^{\infty} \frac{1}{3^{n}} \) is a geometric series with \( r = \frac{1}{3} \). Since \( r < 1 \), this geometric series converges.

Step 4: Apply the Comparison Test

Compare the terms of the given series and the geometric series series: \( \frac{1}{n 3^{n}} \leq \frac{1}{3^{n}} \) for all \( n \geq 1 \). Since the comparison series \( \sum \frac{1}{3^n} \) converges and has greater terms than \( \sum \frac{1}{n 3^n} \), the Comparison Test implies that \( \sum \frac{1}{n 3^n} \) also converges.

Key Concepts

Convergence

In the world of mathematics, the concept of convergence plays a crucial role in understanding infinite series. Convergence occurs when the sum of an infinite series approaches a specific value, rather than just growing indefinitely. If a series converges, we can say that it settles towards a limit. This implies stability in the series, a fundamental property in calculus and analysis.

For instance, consider the series given by \( \sum_{n=1}^{ ightarrow} \frac{1}{3^n} \). Here, each term diminishes as \( n \) increases, making the cumulative sum approach a particular limit. We say this series converges. It is important to understand that convergence is not simply about the series not "growing" but about them reaching or approaching a finite boundary.

Divergence

While convergence refers to reaching a limit, divergence is its counterpart, signifying the lack of a finite sum in the series. A divergent series behaves in such a way that the sum keeps increasing or decreasing indefinitely.

For a series to be divergent, its terms may grow larger as it progresses, or it may oscillate without approaching a particular value. An intuitive way to think of divergence is to picture it like an endless path, constantly extending without settling.

• Divergent series do not satisfy finiteness—that is, they don't stabilize toward a fixed, finite boundary.
• An example of a divergent series would be the harmonic series \( \sum_{n=1}^{ ightarrow} \frac{1}{n} \) which grows indefinitely as \( n \) becomes very large.

Geometric Series

A geometric series is a series where each successive term is obtained by multiplying the previous term by a fixed, constant ratio. This type of series is powerful due to its predictable nature when it comes to convergence.

For a geometric series \( \sum_{n=0}^{ ightarrow} ar^n \) to converge, the common ratio, \( r \), must satisfy \( |r| < 1 \). Otherwise, the series will diverge. Consider our earlier example, \( \sum_{n=1}^{ ightarrow} \frac{1}{3^n} \), which has a ratio \( r = \frac{1}{3} \).

• This specific ratio being less than 1 causes the terms of the series to shrink rapidly, ensuring convergence.
• When \( |r| \ge 1 \), regardless of how large or small the initial term \( a \) is, the series diverges generally.

Series Comparison

The Comparison Test is a useful method for analyzing the behavior of complex series by comparing them with simpler ones whose convergence properties are known. This test allows us to determine whether a given series converges or diverges by relating it to another series.

In order to use the Comparison Test:

• Select a series that is easier to study and has known convergence.
• Ensure that the terms of the given series are compared appropriately with those of the simpler series.
• If the reference series converges and your original series has smaller or equal terms, your series converges as well.
• If the reference series diverges and your original series has larger or equal terms, then the given series diverges too.

For our given series \( \sum_{n=1}^{ ightarrow} \frac{1}{n 3^n} \), we compared it to \( \sum_{n=1}^{ ightarrow} \frac{1}{3^n} \). By observing that \( \frac{1}{n3^n} \leq \frac{1}{3^n} \) and knowing \( \sum_{n=1}^{ ightarrow} \frac{1}{3^nb} \) converges, we concluded that our original series converges by the Comparison Test.