Problem 25 - Calculus: Early Transcendentals

Question

Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum. \( \displaystyle \sum_{n = 1}^{\infty} \frac {e^{2n}}{6^{n - 1}} \)

Step-by-Step Solution

Answer

The series is divergent because \(|r| > 1\).

Step 1: Identify the first term and the common ratio

The given series is \( \sum_{n = 1}^{\infty} \frac {e^{2n}}{6^{n - 1}} \). Start by expressing this in a standard geometric series form. Notice that for \( a = e^2 \) and \( r = \frac{e^2}{6} \), the series can be rewritten as \( \sum_{n = 1}^{\infty} a r^{n-1} \). Here, the first term \( a = e^2 \) and the common ratio \( r = \frac{e^2}{6} \).

Step 2: Calculate the common ratio

Calculate the value of the common ratio \( r = \frac{e^2}{6} \). Check whether \(|r| < 1\) which is the condition for convergence in geometric series.

Step 3: Check for convergence conditions

Compute \(|r|\) to check the convergence. Here, \(|r| = \left| \frac{e^2}{6} \right|\). Since \( e^2 \approx 7.389 \), then \(\frac{e^2}{6} \approx 1.2315\), and \(|r| > 1\).

Step 4: Determine convergence or divergence

Since \(|r| > 1\), the series does not satisfy the convergence condition \(|r| < 1\). Therefore, the series is divergent.

Key Concepts

Convergence and Divergence

Understanding whether a series converges or diverges is crucial in determining its behavior as the number of terms approaches infinity. For a geometric series, the convergence condition can be examined using the absolute value of the common ratio \(|r|\). If \(|r| < 1\), the series converges, meaning that the sum of the infinite series is finite. However, if \(|r| \geq 1\), the series diverges, indicating that the sum grows infinitely and does not settle to a specific value.

For the series in question, we found that the common ratio is \(r = \frac{e^2}{6}\), which approximately equals 1.2315. Since \(1.2315 > 1\), the series diverges, meaning it does not have a finite sum. Recognizing the convergence status of a series guides us on whether it has practical applications or implications in real-world scenarios.

Geometric Series Formula

A geometric series is a series of the form \(a + ar + ar^2 + ar^3 + \ldots\) where \(a\) is the first term and \(r\) is the common ratio. The formula to calculate the sum of an infinite geometric series is given by

• \(S = \frac{a}{1-r}\)

provided that \(|r| < 1\). This formula helps us quickly determine the sum when a series is convergent.

In the given problem, although the structure of the series follows the geometric series pattern, we found that the common ratio \(|r|\) is greater than 1. Consequently, the formula for the sum cannot be applied because the series diverges. Thus, there is no finite sum to calculate.

Common Ratio

The common ratio \(r\) in a geometric series is a critical element as it dictates whether the series will converge or diverge. It is determined by dividing any term in the series by the previous term. In mathematical terms, for consecutive terms \(T_n\) and \(T_{n+1}\), the common ratio \(r\) is \(\frac{T_{n+1}}{T_n}\).

In our series \( \sum_{n = 1}^{\infty} \frac {e^{2n}}{6^{n - 1}} \), we identified \(r\) as \(\frac{e^2}{6}\). This ratio informs us about the scale or rate at which the terms of the series increase or decrease as the series progresses. Since in our case, the common ratio is greater than 1, each term of the series is increasing, leading to divergence.

The common ratio is not just a number; it is a powerful indicator that tells us about the nature and characteristics of the series we are analyzing.