Problem 15 - Calculus: Early Transcendentals

Question

Determine whether the series converges or diverges. \( \displaystyle \sum_{n = 1}^{\infty} \frac {4^{n + 1}}{3^n - 2} \)

Step-by-Step Solution

Answer

The series diverges by the divergence test.

Step 1: Simplify the Series Term

The series given is \( \sum_{n=1}^{\infty} \frac{4^{n+1}}{3^n - 2} \). Start by simplifying the general term of the series, \( a_n = \frac{4^{n+1}}{3^n - 2} \). Notice that \( 4^{n+1} = 4 \cdot 4^n \). Thus, \( a_n = \frac{4 \cdot 4^n}{3^n - 2} \).

Step 2: Test for Dominant Growth

To understand the behavior of \( a_n \), observe that as \( n \to \infty \), both the numerator \( 4^n \) and the denominator \( 3^n - 2 \) are exponential functions. However, the growth of \( 4^n \) is faster than that of \( 3^n \), since the base 4 is larger than the base 3. Therefore, for large \( n \), \( a_n \approx \frac{4 \cdot 4^n}{3^n} = 4 \cdot \left(\frac{4}{3}\right)^n \).

Step 3: Apply the Divergence Test

According to the divergence test, if \( \lim_{n \to \infty} a_n eq 0 \), the series diverges. The expression \( 4 \cdot \left(\frac{4}{3}\right)^n \) is an exponential function of \( n \) that grows without bound (or goes to infinity), as the base \( \frac{4}{3} > 1 \). Thus, \( \lim_{n \to \infty} \left(\frac{4 \cdot 4^n}{3^n - 2}\right) = \infty \).

Step 4: Conclusion on Convergence

Because \( \lim_{n \to \infty} a_n = \infty \), which is not 0, the series diverges by the divergence test. The terms do not approach zero, hence the series cannot converge.

Key Concepts

Divergence Test

The divergence test is a powerful tool in determining if a series converges or diverges.
If you want to check whether a series diverges, the divergence (or nth-term) test provides a simple method to apply.

Here’s how it works:

• Look at the limit of the sequence of terms of the series, specifically: \( \lim_{n \to \infty} a_n \).
• If this limit is not zero, or does not exist, then the series must diverge.

It's important to remember that if the limit does equal zero, the test is inconclusive.
It doesn't necessarily mean the series converges.
Instead, you would need to apply additional tests for convergence.

In our original exercise, the term \( a_n = \frac{4 \cdot 4^n}{3^n - 2} \) leads us directly to apply the test.
Since it doesn't approach zero, the series diverges.

Exponential Functions

Exponential functions play a critical role in determining the behavior of sequences and series.
These functions are of the form \( a^x \) where \( a \) is a constant base.
Because of their rapid growth or decay, they become dominant in many mathematical expressions and problems.

Key characteristics of exponential functions that are crucial include:

• The base \( a \) determines the behavior: if \( a>1 \), the function grows as \( x \to \infty \); if \( 0
• Exponential growth is faster than polynomial growth or linear growth.

In the context of our problem, both the numerator \( 4^n \) and denominator \( 3^n - 2 \) are exponential.
But because 4 is greater than 3, the term \( 4^n \) outpaces \( 3^n \) as \( n \to \infty \), impacting the series' convergence.

Dominant Behavior in Sequences

Understanding which part of a function grows faster is crucial when dealing with sequences and limits.
This concept is prevalent in dealing with exponential terms, where some parts of the expression may "dominate" others.

To analyze this:

• Compare the exponents or bases of exponential terms to see which grows faster.
• The faster-growing term will influence the limit as \( n \to \infty \).

In our series, contrasting bases 4 and 3 show that \( 4^n \) grows quicker than \( 3^n \),
making it a dominant term.
This results in \( a_n \approx \frac{4 \cdot 4^n}{3^n} = 4 \cdot \left(\frac{4}{3}\right)^n \), causing the series to diverge.