Problem 6 - Thomas Calculus

Question

In Exercises \(1-8,\) use the Ratio Test to determine if each series conyerges ahsolutely or diveroes. $$\sum_{n=2}^{\infty} \frac{3^{n+2}}{\ln n}$$

Step-by-Step Solution

Answer

The series diverges by the Ratio Test.

Step 1: Write the General Term of the Series

The general term of the series is given by \( a_n = \frac{3^{n+2}}{\ln n} \).

Step 2: Apply the Ratio Test Formula

According to the Ratio Test, we examine the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If \( L < 1 \), the series converges absolutely; if \( L > 1 \), the series diverges; if \( L = 1 \), the test is inconclusive.

Step 3: Compute the Next Term

The next term \( a_{n+1} \) is \( \frac{3^{(n+1)+2}}{\ln(n+1)} = \frac{3^{n+3}}{\ln(n+1)} \).

Step 4: Set Up the Ratio and Simplify

Calculate the ratio \( \frac{a_{n+1}}{a_n} = \frac{3^{n+3}}{\ln(n+1)} \cdot \frac{\ln n}{3^{n+2}} = \frac{3 \ln n}{\ln(n+1)} \).

Step 5: Find the Limit of the Ratio

Evaluate \( L = \lim_{n \to \infty} \left| \frac{3 \ln n}{\ln(n+1)} \right| \). This simplifies to \( 3 \cdot \lim_{n \to \infty} \frac{\ln n}{\ln(n+1)} \).

Step 6: Simplify the Limit Expression

The internal limit \( \lim_{n \to \infty} \frac{\ln n}{\ln(n+1)} \) simplifies to 1 as both the numerator and the denominator approach infinity and have the same growth rate.

Step 7: Conclusion of the Ratio Test

Therefore, we have \( L = 3 \), which is greater than 1. This means the series diverges according to the Ratio Test.

Key Concepts

Convergence and Divergence

Understanding whether an infinite series converges or diverges is essential for evaluating its behavior. "Convergence" means that as you add more terms of the series, the sum approaches a specific finite value. On the other hand, "divergence" implies that the sum either grows without bound or oscillates indefinitely without settling on a finite value.

The Ratio Test is a common method used to determine the convergence or divergence of a series. Here's how it works:

• Calculate the ratio of successive terms in the series.
• Take the limit of this ratio as the number of terms goes to infinity.
• If the limit is less than 1, the series converges absolutely.
• If the limit is greater than 1, the series diverges.
• If the limit is exactly equal to 1, the test is inconclusive.

In our example exercise, the limit was found to be greater than 1, leading to the conclusion that the series diverges.

Infinite Series

An infinite series is a sum of an infinite sequence of terms. Formally, it is expressed as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the terms of the series. Understanding infinite series is crucial in calculus and mathematical analysis, as they can model several natural phenomena.

There are different types of infinite series:

• Geometric Series: Each term is a constant multiple of the previous term.
• Arithmetic Series: The difference between consecutive terms is constant.
• Harmonic Series: The terms are reciprocals of positive integers.

The series in the original exercise is more complex as it involves an exponential function, \( 3^n \), making it neither purely arithmetic nor geometric. Still, the Ratio Test is a powerful tool in understanding whether such complex series will converge or diverge.

Logarithmic Functions

Logarithmic functions are inverse operations of exponential functions. They play a significant role in mathematics, particularly in calculus, due to their unique properties, such as transforming multiplication into addition. In the problem provided, the natural logarithm \( \ln n \) appears in the denominator of the series' terms, which influences the behavior and convergence of the series.

Here's a quick recap about logarithms:

• The natural logarithm, \( \ln \), has a base of \( e \), where \( e \approx 2.718 \).
• \( \ln(ab) = \ln a + \ln b \)
• \( \ln(a^b) = b \ln a \)
• \( \ln(1) = 0 \)

In the context of our exercise, as \( n \) grows larger, both \( \ln n \) and \( \ln(n+1) \) approach infinity. Despite this, they grow at a similar rate, leading towards simplification when evaluating limits, as seen in our calculation using the Ratio Test.