Problem 40 - Thomas Calculus

Question

Which of the series in Exercises 13 46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{e^{n}}{\left(10+e^{n}\right)^{2}} $$

Step-by-Step Solution

Answer

The series converges by the Limit Comparison Test with \(\sum_{n=1}^{\infty} e^{-n}\).

Step 1: Analyze Dominant Behavior in Denominator

First, consider the behavior of the sequence for large values of \(n\). The term \(e^n\) grows much faster than the constant 10, so for large \(n\), we can approximate \[(10 + e^n)^2 \approx (e^n)^2 = e^{2n}\]This simplification will help us compare the original series to a known series.

Step 2: Simplify the Term

By using the approximation from Step 1, the term of the series can be simplified as follows:\[\frac{e^n}{(10 + e^n)^2} \approx \frac{e^n}{e^{2n}} = e^{-n}\]This transformed term is key to determining convergence.

Step 3: Compare to a Known Convergent Series

The series \(\sum_{n=1}^{\infty} e^{-n}\) is a geometric series with ratio \(r = \frac{1}{e} < 1\). A geometric series converges if \(|r| < 1\). Hence, \(\sum_{n=1}^{\infty} e^{-n}\) converges.

Step 4: Apply the Limit Comparison Test

Use the Limit Comparison Test with the series \(\sum_{n=1}^{\infty} e^{-n}\). Calculate the following limit:\[\lim_{{n \to \infty}} \frac{\frac{e^n}{(10 + e^n)^2}}{e^{-n}} = \lim_{{n \to \infty}} \frac{e^n \cdot e^n}{(10 + e^n)^2} = \lim_{{n \to \infty}} \frac{e^{2n}}{(10 + e^n)^2} = \lim_{{n \to \infty}} \frac{e^{2n}}{e^{2n}} = 1\]Since the limit is finite and positive, by the limit comparison test, the original series converges because \(\sum_{n=1}^{\infty} e^{-n}\) converges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a powerful tool to evaluate the convergence or divergence of a series. It simplifies the analysis by letting us compare a complicated series to a simpler, well-known one. To use this test, we need two series: the targeted series \, \(\sum a_n\) \, and a comparison series \, \(\sum b_n\), which is either known to converge or diverge.

Here's how it works in steps:

• First, identify a comparison series \, \(\sum b_n\), similar in form to \, \(\sum a_n\)\.
• Next, compute the limit \(\lim_{{n \to \infty}} \frac{a_n}{b_n}\).
• If this limit is a positive, finite number, both series will either converge or diverge together.

In the exercise, the original series, \(\sum \frac{e^n}{(10+e^n)^2}\), was compared to the geometric series \(\sum e^{-n}\). The calculated limit was \(1\), indicating that both series share the same convergence behavior.

Using the Limit Comparison Test allows complex series to be examined more easily using the known properties of simpler series.

Geometric Series

A geometric series is one of the most straightforward series types to understand. It has the general form \(\sum_{n=0}^{\infty} ar^n\), where \(a\) is the first term and \(r\) is the common ratio. The key characteristic of a geometric series is its simple structure, which makes it easy to analyze and apply.

Here's what you need to know about geometric series:

• If the absolute value of the common ratio \(|r| < 1\), the series converges.
• If \(|r| \geq 1\), the series diverges.
• The sum of a convergent geometric series is calculated as \(\frac{a}{1-r}\).

In our example, we dealt with a series similar to a geometric one: \(\sum e^{-n}\), which had a common ratio \(r = \frac{1}{e}\), satisfying \(|r| < 1\). As such, it converges.

Recognizing a series as geometric often simplifies your work, as it directly relates to known rules of convergence.

Dominant Behavior Analysis

Dominant Behavior Analysis is a technique used to simplify rational functions, especially when dealing with infinite series. The focus is on identifying the term that grows the fastest, thus driving the behavior of the sequence as \(n\) approaches infinity.

Follow these steps to apply dominant behavior analysis effectively:

• Look at each term in the sequence as \(n\) becomes very large.
• Identify which part of the term grows the fastest and becomes dominant.
• Use this dominant behavior to simplify the series for comparison with known series.

In our analysis, we noted that within the series \(\sum \frac{e^n}{(10 + e^n)^2}\), the term \(e^n\) dominates over the constant \(10\) for large \(n\). Thus, \((10 + e^n)^2\) was approximated by \(e^{2n}\).

Recognizing dominant behavior helps in approximating complex terms, simplifying the comparison between series.

Convergence/Divergence Tests

Mathematicians often use a variety of tests to determine the convergence or divergence of series. These tests provide structured approaches to predict whether a series sums to a finite number or grows indefinitely. Key convergence tests include:

• Comparison Test: Compares a series to another with known convergence behavior.
• Limit Comparison Test: Similar to the comparison test but uses limits for effectiveness in challenging cases.
• Ratio Test: Uses the limit of ratios of successive terms to predict convergence.
• Integral Test: Involves integration to compare a series to a continuous function.
• Geometric Series Test: Specifically for geometric series using the common ratio \(r\).

Each of these tests has distinct criteria and applicability, so selecting the right one depends on the series in question.

In the task at hand, the Limit Comparison Test was used due to its relevancy and effectiveness in dealing with the original series in comparison to the geometric series \(\sum e^{-n}\). By understanding these tests, you can choose the most efficient path to analyze complex series.