Question

Use the integral test to determine whether the following sums converge. \(\sum_{n=1}^{\infty} \frac{e^{n}}{1+e^{2 n}}\)

Short answer

The series diverges because the corresponding integral diverges.

Step-by-step solution

Identify the Function for the Integral Test

To apply the integral test, recognize the sequence for the series as a function. The series is given as \( \sum_{n=1}^{\infty} \frac{e^{n}}{1+e^{2n}} \). Thus, define the function \( f(x) = \frac{e^{x}}{1+e^{2x}} \) which must be continuous, positive, and decreasing for all \( x \geq 1 \).

Check for Positivity and Continuity

Examine the function \( f(x) = \frac{e^{x}}{1+e^{2x}} \). For \( x \geq 1 \), both the numerator \( e^x \) and the denominator \( 1+e^{2x} \) are positive, ensuring \( f(x) > 0 \). The function is also continuous for \( x \geq 1 \) because it is a ratio of continuous functions where the denominator is never zero.

Verify the Function is Decreasing

To ensure that \( f(x) \) is decreasing, check the first derivative, \( f'(x) \):\[ f'(x) = \frac{d}{dx} \left( \frac{e^{x}}{1+e^{2x}} \right) = \frac{e^x(1+e^{2x}) - e^x \cdot 2e^{2x}}{(1+e^{2x})^2} = \frac{e^x(1 + e^{2x} - 2e^{2x})}{(1+e^{2x})^2}\]\[ = \frac{e^x(1 - e^{2x})}{(1+e^{2x})^2} \]The numerator \( e^x(1 - e^{2x}) \) is negative for \( x > 0 \) as \( e^{2x} > 1 \), so \( f'(x) < 0 \) indicating that \( f(x) \) is decreasing for \( x \geq 1 \).

Set up the Improper Integral

The integral test states that the series \( \sum_{n=1}^{\infty} f(n) \) converges if and only if the integral \( \int_{1}^{\infty} f(x) \, dx \) converges. Therefore, consider the improper integral:\[ \int_{1}^{\infty} \frac{e^x}{1+e^{2x}} \, dx \]

Evaluate the Improper Integral

Perform substitution to evaluate the integral. Let \( u = e^{2x} \), resulting in \( du = 2e^{2x} \, dx \). Therefore:\[ dx = \frac{du}{2u} \]Transform the integral:\[ \int \frac{e^x}{1+e^{2x}} \, dx = \int \frac{e^x}{1+u} \cdot \frac{du}{2u} = \frac{1}{2} \int \frac{1}{(1+u)\sqrt{u}} \, du \]This integral does not converge because as \( u \to \infty \), the factor \( \frac{1}{(1+u)\sqrt{u}} \to 0 \) insufficiently rapidly to produce a finite result.

Key concepts

Convergence of series

In mathematics, determining the convergence of a series is crucial to understand whether the sum of its terms will settle into a specific number or diverge endlessly. When working with an infinite series like \( \sum_{n=1}^{\infty} \frac{e^{n}}{1+e^{2n}} \), the question arises: does the sum of all these infinitely many terms lead to a finite number?

The significance of convergence lies in the potential application of such series in real-world problems and mathematical contexts. If a series converges, it means that by adding more terms, we approach a stable sum, allowing for practical calculations and interpretations. Conversely, if it diverges, the series grows without bounds, and its sum cannot be nailed down to a single number.

One of the reliable methods for testing series convergence is the **Integral Test**. This test connects the series to an improper integral and checks if this related integral converges. If it does, so does the series. If not, the series diverges as well. Thus, before applying the integral test, ensuring that the function associated with the series is continuous, positive, and decreasing for all values greater than or equal to 1 is vital. This laid a foundation upon which convergence analysis is performed.

Improper integral evaluation

Evaluating improper integrals is a fascinating yet challenging mathematical adventure. To employ the integral test for a series, we translate the problem into evaluating an improper integral. For instance, to examine the series \( \sum_{n=1}^{\infty} \frac{e^{n}}{1+e^{2n}} \), we define the corresponding function, \(\ f(x) = \frac{e^{x}}{1+e^{2x}} \), and set up the integral \( \int_{1}^{\infty} \frac{e^x}{1+e^{2x}} \, dx \).

An improper integral takes place over an infinite interval, requiring specific techniques for evaluation. Substitution is often a tool of choice. In our case, by letting \( u = e^{2x} \), we can simplify the expression; however, it ultimately doesn't yield a finite result.

When performing these evaluations, special care must be given to the behavior of the integral at infinity. The key question is whether the integral results in a finite value as the upper bound approaches infinity. Unfortunately, in this example, the factor \( \frac{1}{(1+u)\sqrt{u}} \) doesn't shrink fast enough to make the integral converge, indicating that the series does not converge.

Decreasing function verification

A crucial part of applying the integral test is verifying that the function associated with the series is indeed decreasing. For the series \( \sum_{n=1}^{\infty} \frac{e^{n}}{1+e^{2n}} \), we're interested in the function \(\ f(x) = \frac{e^{x}}{1+e^{2x}} \).

To confirm it's decreasing, we calculate its first derivative, \( f'(x) \).
For our function, \( f'(x) = \frac{e^x(1 - e^{2x})}{(1+e^{2x})^2} \). By analyzing this expression, we note that for \(x \geq 1\), the term \(e^{2x}\) exceeds 1, making \(1 - e^{2x} < 0\). Consequently, \( f'(x) < 0 \), confirming \(f(x)\) is decreasing.

This verification establishes that the function is suitable for the integral test, meeting the necessary criteria of being continuous, positive, and decreasing over the specified interval. Understanding this behavior is essential, as the integral test relies on these properties to judge whether the improper integral, and hence the series, converges or diverges.