Problem 20 - Single Variable Calculus

Question

Show that the series diverges. \(\sum_{n=0}^{\infty} \frac{(-1)^{n} 3^{n}}{2^{n-1}}\)

Step-by-Step Solution

Answer

The series \( a_n = \frac{(-1)^{n} 3^{n}}{2^{n-1}} \) diverges because applying the Absolute Convergence Test, we find that the sum of the absolute values of the terms diverges: \( \sum_{n=0}^{\infty} |a_n| = \sum_{n=0}^{\infty} \frac{3^n}{2^{n-1}} \). Using the Ratio Test, we get \( \lim_{n \to \infty} \frac{3}{2} > 1\); thus, the original series diverges.

Step 1: Write down the series

The function can be represented as: \( a_n = \frac{(-1)^{n} 3^{n}}{2^{n-1}} \)

Step 2: Check if the series converges absolutely

The Absolute Convergence Test implies that if the sum of the absolute values of the terms of a series converges, then the original series converges. First, we need to find the absolute value of the terms in the series, which is given by: \( |a_n| = \frac{3^n}{2^{n-1}} \) Now, let's check if the sum of the absolute values converges: \( \sum_{n=0}^{\infty} |a_n| = \sum_{n=0}^{\infty} \frac{3^n}{2^{n-1}} \)

Step 3: Apply the Ratio Test to the sum of absolute values

To determine if the sum of absolute values converges, we can use the Ratio Test. The Ratio Test compares the ratio of consecutive terms in a series: \( \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = L \) If \(L<1\), the series converges; if \(L>1\), the series diverges; if \(L=1\), the test is inconclusive. Compute the limit using the absolute values of the terms: \( \lim_{n \to \infty} \left|\frac{ \frac{3^{n+1}}{2^{n}}}{ \frac{3^n}{2^{n-1}}}\right| = \lim_{n \to \infty} \frac{3^{n+1}2^{n-1}}{3^n2^{n}} \) Simplify the expression: \( \lim_{n \to \infty} \frac{3}{2} \) The limit is \( \frac{3}{2} (= 1.5) > 1 \), which implies that the sum of the absolute values of the terms diverges. According to the Absolute Convergence Test, if the sum of the absolute values of the terms diverges, the original series also diverges. Therefore, we conclude that the series \( \sum_{n=0}^{\infty} \frac{(-1)^{n} 3^{n}}{2^{n-1}} \) diverges.

Key Concepts

Absolute Convergence Test

Understanding the Absolute Convergence Test is crucial when dealing with infinite series. Essentially, this test helps to determine whether a series will converge based on the convergence of its absolute values. For a series \( \sum a_n \), we look at the series formed by the absolute values of the terms, which is \( \sum |a_n| \).

If \( \sum |a_n| \) converges, then we can say that the original series \( \sum a_n \) converges absolutely. This indicates a stronger form of convergence, meaning that not only does the series converge, but it does so in a way that is independent of the order of the terms.

An Important Implication

• If \( \sum |a_n| \) diverges, then \( \sum a_n \) also diverges or at best, converges conditionally (which is a weaker form of convergence and highly dependent on the order of terms).

Thus, when faced with an alternating series such as our exercise \( \sum_{n=0}^{\infty} \frac{(-1)^{n} 3^{n}}{2^{n-1}} \), the first step is to check if the series of absolute terms converges. If it does not, as shown in the exercise when the series of absolute terms diverges, there's no need to look any further—the series does not converge absolutely, meaning it diverges or only converges conditionally.

Ratio Test

The Ratio Test is a popular tool used to assess the convergence of an infinite series. To apply the Ratio Test, you calculate the limit of the absolute value of the ratio of consecutive terms, namely \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

The outcomes of the Ratio Test are pretty straightforward:

• If \( L < 1 \), the series converges.
• If \( L > 1 \), the series diverges.
• If \( L = 1 \), the test is inconclusive, and we need to use other convergence tests.

When we applied the Ratio Test to the absolute series \( \sum |a_n| \) in our exercise, we found that \( L = \frac{3}{2} \) which is greater than 1. This clearly indicates that the series diverges according to the Ratio Test.

Interpreting the Result

It's crucial to interpret the result correctly; a series with \( L > 1 \) does not just fail to converge—it actively diverges. This means that the partial sums of the series will grow without bound as we add more and more terms. The series in our exercise, therefore, does not sum to a finite value.

Series Convergence

Series convergence is a fundamental concept in calculus and mathematical analysis. It refers to the behavior of an infinite series \( \sum a_n \), which is an ordered sum of terms. An infinite series converges if the sequence of its partial sums \( S_N = a_1 + a_2 + \cdots + a_N \) approaches a limit as \( N \), the number of terms, becomes large.

For a series to be convergent, adding more terms to the sum should bring the total sum closer to a specific finite value, rather than making it grow indefinitely or oscillate without settling down.

Different Types of Convergence

There are two primary types of convergence for series - absolute convergence and conditional convergence.

• Absolute Convergence: A series is said to be absolutely convergent if the series of the absolute values of its terms converges.
• Conditional Convergence: This occurs when a series converges but does not converge absolutely.

In our exercise, we showed that the series diverges because the series of absolute values does not converge. This is an essential part of series analysis, as the behavior of the series can significantly impact the outcome of problems in various fields, including engineering, physics, and economics.