Problem 42 - Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The series is divergent.

Step 1: Understanding the Series

The series given is \( \sum_{n=1}^{\infty} \frac{e^n}{n^2} \). We need to determine if this series converges or diverges.

Step 2: Apply the Ratio Test

The Ratio Test is a method to determine the convergence of a series. For a series \( \sum a_n \), compute \( L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} \). If \( L > 1 \) or \( L = \infty \), the series diverges; if \( L < 1 \), it converges.In our series, \( a_n = \frac{e^n}{n^2} \). Let's calculate:\[ \frac{a_{n+1}}{a_n} = \frac{\frac{e^{n+1}}{(n+1)^2}}{\frac{e^n}{n^2}} = \frac{e \cdot n^2}{(n+1)^2} \]Simplifying, we have:\[ \frac{e \cdot n^2}{(n+1)^2} \to \frac{e n^2}{n^2} = e \left( \frac{n^2}{(n+1)^2} \right) \]

Step 3: Calculate the Limit

As \( n \to \infty \), \( \frac{n^2}{(n+1)^2} = \left( \frac{n}{n+1} \right)^2 \to 1 \). Thus, our expression becomes: \( e \cdot 1 = e \).So, \( L = e \). Since \( e \approx 2.718 > 1 \), the Ratio Test indicates the series is divergent.

Step 4: Conclusion

Since the limit \( L = e \) is greater than 1, according to the Ratio Test, the series \( \sum_{n=1}^{\infty} \frac{e^n}{n^2} \) is divergent.

Key Concepts

Ratio Test

The Ratio Test is a crucial technique used in determining the behavior of infinite series. It helps decide if a series is convergent or divergent based on the limit of the ratio of consecutive terms. To apply the Ratio Test, we examine a series of the form \( \sum a_n \). The process involves calculating the limit \( L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} \).

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \) or \( L = \infty \), the series diverges.

In this specific exercise, the series \( \sum_{n=1}^{\infty} \frac{e^n}{n^2} \) was tested using the Ratio Test. By calculating the limit as \( n \to \infty \) for the ratio \( \frac{a_{n+1}}{a_n} \), where \( a_n = \frac{e^n}{n^2} \), we found that \( L = e \). Since \( e \approx 2.718 \) and is greater than 1, the Ratio Test indicates that the series is divergent.

Divergence

In the context of series, divergence means that the series does not settle at a particular value. Simply put, as you keep adding more terms to the series, it grows without bound or oscillates indefinitely.For the series \( \sum_{n=1}^{\infty} \frac{e^n}{n^2} \), divergence was confirmed through the Ratio Test. This tells us that as \( n \) grows large, the terms do not shrink quickly enough to result in convergence.Understanding divergence is essential because it highlights that certain series will not yield a finite sum. In the exercise, because the limit \( L = e > 1 \) when applying the Ratio Test, the conclusion was that the infinite addition of terms in this series grows steadily, without ever reducing to a single sum.

Limit Calculation

When analyzing series and convergence, calculating limits is often necessary to apply tests like the Ratio Test. It involves determining what happens to a sequence or expression as the term number \( n \) increases indefinitely (\( n \to \infty \)).In this problem, the critical limit calculation was for the expression \( \lim_{n \to \infty} \frac{e \cdot n^2}{(n+1)^2} \). By simplifying, we ended up with \( \lim_{n \to \infty} e \left( \frac{n}{n+1} \right)^2 \), which approaches \( e \times 1 = e \) because \( \left( \frac{n}{n+1} \right)^2 \to 1 \) as \( n \to \infty \).Recognizing how to compute these kinds of limits enables a deeper understanding of how series behave when extended infinitely.

Infinite Series

An infinite series is essentially the sum of infinitely many terms, expressed as \( \sum_{n=1}^{\infty} a_n \). The goal often is to determine whether such a sum converges to a finite value or diverges.Infinite series are significant in various fields, including mathematics and physics, because they can model continuous processes or distributions. They form an essential part of analysis, finding applications in calculating things like probabilities, series expansions, and in various algorithms.In this exercise, the series \( \sum_{n=1}^{\infty} \frac{e^n}{n^2} \) was under consideration. We applied the Ratio Test to assess convergence, concluding it diverged. This insight into infinite series broadens the understanding of mathematical functions and growth, showcasing why knowing their properties is vital.