Problem 6 - Calculus: Early Transcendentals

Question

Use the Integral Test to determine whether the series is convergent or divergent. \( \displaystyle \sum_{n = 1}^{\infty} \frac {1}{\left(3n - 1 \right)^34} \)

Step-by-Step Solution

Answer

The series converges.

Step 1: Identify the General Term

The given series is \( \sum_{n=1}^{\infty} \frac{1}{(3n-1)^3} \). The general term of this series is \( a_n = \frac{1}{(3n-1)^3} \).

Step 2: Define the Related Function, f(x)

For the integral test, identify a related continuous, positive, and decreasing function \( f(x) \). A natural choice is \( f(x) = \frac{1}{(3x-1)^3} \).

Step 3: Check if f(x) is Positive and Decreasing

For \( x \geq 1 \), \( (3x-1)^3 > 0 \), so \( f(x) > 0 \). To show \( f(x) \) is decreasing, confirm that its derivative \( f'(x) \) is negative for \( x \geq 1 \). Calculate \( f'(x) = -\frac{9}{(3x-1)^4} \), which is negative for all \( x \geq 1 \). Thus, \( f(x) \) is decreasing.

Step 4: Set Up the Integral

Set up the improper integral \( \int_{1}^{\infty} \frac{1}{(3x-1)^3} \, dx \) to test for convergence.

Step 5: Substitute and Simplify

Let \( u = 3x-1 \), then \( du = 3 \, dx \) or \( dx = \frac{du}{3} \). Change the limits: when \( x=1 \), \( u=2 \); when \( x=\infty \), \( u=\infty \). The integral becomes \( \frac{1}{3} \int_{2}^{\infty} \frac{1}{u^3} \, du \).

Step 6: Evaluate the Integral

Evaluate \( \frac{1}{3} \int_{2}^{\infty} \frac{1}{u^3} \, du = \frac{1}{3} \left[ -\frac{1}{2u^2} \right]_{2}^{\infty} = \frac{1}{3} \left( 0 - \left(-\frac{1}{8} \right) \right) = \frac{1}{24} .\)

Step 7: Conclude Based on the Integral Test

Since the improper integral \( \int_{1}^{\infty} \frac{1}{(3x-1)^3} \, dx \) converges to \( \frac{1}{24} \), the original series \( \sum_{n=1}^{\infty} \frac{1}{(3n-1)^3} \) is also convergent by the Integral Test.

Key Concepts

Series Convergence

When you come across a series, the first question that arises is whether it converges or diverges. Convergence means that as you add more and more terms, the series approaches a specific number. Divergence, on the other hand, suggests that the series keeps growing indefinitely or fluctuates without settling.
To determine convergence, several tests can be used, such as the Integral Test, which is especially useful for series that resemble integrals. Convergence is essential not only in mathematics but also in applications like calculating areas, predicting behaviors, and more.

Improper Integral

An improper integral extends the concept of a definite integral to cases where the interval is infinite or the function becomes unbounded. For instance, the interval \([a, \, \infty)\) requires evaluation of limits to determine the integral's convergence.
This concept is critical since improper integrals help in understanding series, particularly when using the Integral Test. To evaluate an improper integral, we set boundaries and calculate limits to assess if the integral converges to a finite value.
In our example, the task was to solve \( \int_{1}^{\infty} \frac{1}{(3x-1)^3} \, dx \). The integral's success in producing a finite value indicates convergence.

Decreasing Function

A decreasing function steadily declines in value as the input increases. In the context of the Integral Test, the related function must be continuous, positive, and decreasing from a certain point onward. This ensures the function's behavior is stable enough to predict the series' convergence.
To verify a decreasing function, examine its derivative. A negative derivative across the interval confirms the function is falling. For example, in \( f(x) = \frac{1}{(3x-1)^3} \), calculating the derivative gives \( f'(x) = -\frac{9}{(3x-1)^4} \), which is negative for \( x \geq 1 \). This supports that \( f(x) \) is indeed decreasing, qualifying it for the Integral Test.

Substitution Method

The substitution method simplifies complex integrals by transforming the integral into a more manageable form. You set a variable (often \( u \)) to replace a function part, making the integration process simpler.
In tackling \( \int_{1}^{\infty} \frac{1}{(3x-1)^3} \, dx \), substitution is key. By letting \( u = 3x-1 \), and \( du = 3 \, dx \), we transformed the integral limits and form. By doing this, it converted the integral into \( \frac{1}{3} \int_{2}^{\infty} \frac{1}{u^3} \, du \), simplifying evaluation.
This approach reduces complexity, ensuring that even integrals that initially appear challenging can be solved with fundamental techniques.