Problem 5 - Thomas Calculus

Question

Use the Integral Test to determine if the series in Exercises \(1-10\) converge or diverge. Be sure to check that the conditions of the Integral Test are satisfied. $$ \sum_{n=1}^{\infty} e^{-2 n} $$

Step-by-Step Solution

Answer

The series converges.

Step 1: Check the Function and Conditions

First, consider the function \( f(n) = e^{-2n} \). To apply the Integral Test, we need to ensure that \( f(x) = e^{-2x} \) is positive, continuous, and decreasing for \( x \geq 1 \). Since the exponential function \( e^{-2x} \) is positive for all \( x \), continuous, and its derivative \( f'(x) = -2e^{-2x} \) is negative, \( f(x) \) is decreasing for \( x \geq 1 \). Thus, the conditions for the Integral Test are satisfied.

Step 2: Set up the Integral

Next, set up the improper integral \( \int_{1}^{\infty} e^{-2x} \, dx \) to evaluate the convergence of the series \( \sum_{n=1}^{\infty} e^{-2n} \).

Step 3: Integrate the Function

Compute the integral: \[\int e^{-2x} \, dx = -\frac{1}{2}e^{-2x} + C.\] Now evaluate it from 1 to \( \infty \): \[\int_{1}^{\infty} e^{-2x} \, dx = \left[-\frac{1}{2}e^{-2x}\right]_{1}^{\infty}.\]

Step 4: Evaluate the Limits

Evaluate the integral's limits: - As \( x \) approaches infinity, \( e^{-2x} \) approaches 0, so \(-\frac{1}{2}e^{-2x} \to 0\).- When \( x = 1 \), \(-\frac{1}{2}e^{-2\cdot1} = -\frac{1}{2}e^{-2}\).Putting these together, \[\int_{1}^{\infty} e^{-2x} \, dx = 0 - \left(-\frac{1}{2}e^{-2}\right) = \frac{1}{2e^2}. \]

Step 5: Conclude with the Integral Test

Since the improper integral \( \int_{1}^{\infty} e^{-2x} \, dx \) converges to \( \frac{1}{2e^2} \), by the Integral Test, the series \( \sum_{n=1}^{\infty} e^{-2n} \) also converges.

Key Concepts

Convergence of Series

Series are like long sums that add infinite terms together. We often need to know if these sums ever settle on a specific number or just keep getting bigger and bigger. This is where the concept of convergence comes in. A series is said to converge if this infinite sum approaches a particular number as you keep adding more terms. To study convergence, mathematicians often use tests. One such test is the **Integral Test**, which is ideal for series where each term comes from a function that can be integrated. With the Integral Test, if the integral of the function over a certain range converges, then the series converges too. This test is particularly useful for series that involve exponential, polynomial, or logarithmic functions. - **Conditions for the Integral Test**: - The function must be positive. - It should be continuous. - It must decrease over the range we are looking at. When these conditions are met, testing the series for convergence becomes much easier.

Improper Integrals

Improper integrals solve a special kind of problem when determining the area under a curve that stretches towards infinity. You can't simply calculate them like a standard integral because one or more limits of integration are infinite. Such integrals also appear when a function has discontinuities or spikes.For the series in our exercise, the improper integral helps us understand if the infinite sum of terms approaches a number. When we evaluate these integrals, we do so by taking limits. Take, for example, \[ \int_{1}^{b} e^{-2x} \, dx \]as \(b\) approaches infinity. If we can find a finite number to which this integral approaches, the integral converges, suggesting that the related series does too.**Steps to Evaluate**:- Replace the infinity symbol with a variable, such as \(b\).- Evaluate the integral from 1 to \(b\).- Compute the limit as \(b\) heads towards infinity.If this value exists, the integral and thus the original series converge.

Exponential Functions

Exponential functions are functions like \(e^{-2x}\) that include a constant raised to the power of a variable. They grow (or shrink) very rapidly. The base 'e' is a special number in mathematics, approximately 2.718, that plays a crucial role in growth and decay models.Why exponential functions are important:- They model real-world phenomena like radioactive decay, population growth, and interest rates.- They appear frequently in calculus and differential equations.- They are one of the simplest kinds of functions that describe compounding processes, hence, their frequent presence in convergence tests.For our series, \(e^{-2n}\) denotes an exponential decay. As \(n\) gets larger, \(e^{-2n}\) gets very small very quickly. This rapid decrease is why its sum may converge. In calculus, we love exponential functions for their neat properties. They are straightforward to differentiate and integrate, making them perfect for studying the convergence of series.