Problem 37 - 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.) $$ \sum_{n=1}^{\infty} \frac{e^{n}}{1+e^{2 n}} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{e^n}{1+e^{2n}} \) converges by the Limit Comparison Test.

Step 1: Identify the type of series

The series presented is \( \sum_{n=1}^{\infty} \frac{e^n}{1+e^{2n}} \). This is an infinite series where each term is \( \frac{e^n}{1+e^{2n}} \). Our task is to determine whether this series converges or diverges.

Step 2: Simplify the series terms

Examine the term \( a_n = \frac{e^n}{1+e^{2n}} \). As \( n \) increases, \( e^{2n} \) becomes very large. In such a case, \( 1+e^{2n} \approx e^{2n} \), so the term \( a_n \approx \frac{e^n}{e^{2n}} = \frac{1}{e^n} \).

Step 3: Use a convergence test

Apply the Limit Comparison Test with \( b_n = \frac{1}{e^n} \), which is a geometric series with a common ratio of \( \frac{1}{e} \) where \( e > 1 \). We know the geometric series \( \sum \frac{1}{e^n} \) converges because the common ratio is less than 1.

Step 4: Perform the Limit Comparison Test

Calculate \( \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{e^n}{1+e^{2n}}}{\frac{1}{e^n}} = \lim_{n \to \infty} \frac{e^{2n}}{1+e^{2n}} \). Simplifying further, we get \( \lim_{n \to \infty} \frac{1}{1+\frac{1}{e^{2n}}} = 1 \). Since this limit is a finite positive number, the Limit Comparison Test shows that \( \sum_{n=1}^{\infty} \frac{1}{e^n} \) and \( \sum_{n=1}^{\infty} \frac{e^n}{1+e^{2n}} \) behave similarly.

Step 5: Conclusion

According to the Limit Comparison Test, since \( \sum \frac{1}{e^n} \) converges, and the limit computed is positive and finite, the given series \( \sum \frac{e^n}{1+e^{2n}} \) also converges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a vital tool when determining the convergence or divergence of an infinite series. This test involves comparing a given series with another series, whose convergence behavior is well-known. The key is to find a series, say \( b_n \), to compare to the original series \( a_n \). For the test, you compute:

• \( \lim_{n \to \infty} \frac{a_n}{b_n} \)

This limit should yield a finite, positive number. If it does, both series \( \sum a_n \) and \( \sum b_n \) will either both converge or both diverge.
This method simplifies complex series evaluation because it relates the unknown behavior of \( \sum a_n \) to the known behavior of \( \sum b_n \). In practice, finding a suitable \( b_n \) often involves simplifying the terms \( a_n \) for large \( n \), as seen in our example.

Infinite Series

An infinite series is essentially the sum of infinitely many terms. Mathematically, it can be expressed as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents each term of the sequence.
Analyzing infinite series revolves around understanding their convergence: can they settle into a stable total sum as the number of terms increases infinitely?

• If the sum approaches a specific finite number, the series converges.
• If it grows indefinitely or oscillates without settling, the series diverges.

Infinite series are integral to calculus and analytical mathematics, offering insight into functions, power series expansions, and more. However, determining convergence can be complex, necessitating various strategies including the comparison tests, ratio test, and more.

Geometric Series

A geometric series is a specific type of series where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. The general form is \( a, ar, ar^2, ar^3, \dots \), which can be summed as \( \sum_{n=0}^{\infty} ar^n \).
The convergence or divergence of a geometric series is simple to determine:

• If the common ratio \( |r| < 1 \), the series converges.
• If \( |r| \geq 1 \), the series diverges.

The geometric series \( \sum \frac{1}{e^n} \), used in our example, has a convergent nature because its common ratio \( r = \frac{1}{e} < 1 \). This characteristic makes it an ideal candidate for comparison in convergence tests.

Convergence and Divergence

Convergence and divergence are fundamental concepts when studying infinite series. A series is said to converge if the sum of its terms approaches a finite number as more and more terms are added. Conversely, it diverges if the sum does not settle to a single value.

• **Convergent series:** These series have finite sums. Techniques like the limit comparison test can show whether a series behaves like known convergent series.
• **Divergent series:** These lack a defined, finite sum. They may increase indefinitely or simply fail to settle.

Different tests are employed to assess convergence or divergence, each leveraging unique properties of sequences and series. Recognizing these behaviors is crucial in mathematical analysis, finding applications across pure and applied mathematics.