Problem 13 - Essential Calculus Early Transcendentals

Question

Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum_{n=1}^{\infty} \frac{3^{n}}{e^{n-1}}$$

Step-by-Step Solution

Answer

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

Step 1: Rewrite the Series

To begin analyzing the convergence of the series, rewrite the series in a simplified form. The given series is \[ \sum_{n=1}^{\infty} \frac{3^{n}}{e^{n-1}}. \] This can be rewritten as \[ \sum_{n=1}^{\infty} \frac{3^{n}}{e^{n}} \cdot e. \]Simplifying further, we get:\[ e \sum_{n=1}^{\infty} \left( \frac{3}{e} \right)^{n}. \] Now, it is in the form suitable for recognizing it as a geometric series.

Step 2: Identify the Geometric Series

The series is now:\[ e \sum_{n=1}^{\infty} \left( \frac{3}{e} \right)^{n}. \] This is a geometric series with the common ratio \( r = \frac{3}{e} \). A geometric series of the form \( \sum_{n=0}^{\infty} ar^n \) converges if \( |r| < 1 \) and diverges otherwise.

Step 3: Check Convergence Conditions

Check the condition for convergence of a geometric series, \(|r| < 1\). Here, \( r = \frac{3}{e} \), so calculate \(|r| = \left| \frac{3}{e} \right| \approx 1.103\). Since \(|r| > 1\), the series does not satisfy the condition for convergence.

Step 4: Conclusion on Convergence

Since \(|r| \approx 1.103 > 1\), the series diverges. Therefore, we cannot find a sum for this series as it does not converge.

Key Concepts

Series Convergence

Understanding series convergence is essential in calculus. When we have an infinite series, we need to determine whether it sums to a finite number or not. This is what we call convergence. If a series converges, it means adding up all its terms results in a finite value. Otherwise, the series is divergent and does not settle on any particular sum.

For example, consider an infinite series like \[ \sum_{n=1}^{\infty} a_n. \]To test for convergence, we often need to apply specific tests that simplify the terms in the series, like the geometric series test.

It's important to know that various types of series have different characteristics that influence convergence. Understanding these can greatly ease tackling any calculus problem involving series.

Geometric Series Ratio

A fundamental idea in understanding geometric series is the notion of the geometric series ratio. Every geometric series takes the form \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term and \( r \) is the common ratio.

The geometric series ratio, \( r \), determines the behavior of the series. By checking its absolute value, we can immediately predict whether the series converges or diverges.

- If \( |r| < 1 \), the series converges.- Otherwise, if \( |r| \geq 1 \), the series diverges.
In our exercise, the ratio \( r = \frac{3}{e} \) is approximately 1.103, meaning \( |r| > 1 \). This condition leads directly to the conclusion that the series cannot converge, highlighting the importance of accurately determining the common ratio.

Geometric Series Test

The geometric series test is a quick method to evaluate the convergence of geometric series. Its simplicity stems from focusing solely on the series' common ratio.

To apply this test, you:

• Identify the series as geometric.
• Determine the common ratio \( r \).
• Evaluate \( |r| \).

If \( |r| < 1 \), the series converges, and we can even calculate its sum using the formula:\[ S = \frac{a}{1-r} \]if the series has non-zero first term \( a \). However, if \( |r| \geq 1 \), the series diverges, so no sum exists.

In our problem, the ratio \( |r| = |\frac{3}{e}| \) was greater than 1, simply showing the series diverges. This test shows how powerful a single calculation can be in determining a series' fate.

Calculus Divergence

Calculus divergence refers to the behavior of an infinite series when it does not settle on a finite sum. This is when the series is termed divergent. Divergence indicates that as you add more terms in the series, the total just grows without bound or fluctuates endlessly.

Divergence is common in series and can result from having common ratios with absolute values equal to or greater than 1 in geometric series. These conditions prevent series from converging into any particular value.

For the given exercise, our series diverged as the common ratio \( |\frac{3}{e}| \) was found to be greater than 1. Here, practicing the series convergence tests and understanding diverging conditions ensures we gain insights into the nature of sum behaviors. Recognizing divergence is crucial, as it's a fundamental part of building on complex calculus concepts.