Problem 6 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The series diverges because the limit in the Ratio Test is 3, which is greater than 1.

Step 1: Write the General Term

The general term of the series is given by \( a_n = \frac{3^{n+2}}{\ln n} \). For the Ratio Test, we will use \( a_n = 3^2 \cdot \frac{3^n}{\ln n} = \frac{9 \cdot 3^n}{\ln n} \).

Step 2: Apply the Ratio Test Formula

The Ratio Test involves computing \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Here, \( a_{n+1} = \frac{9 \cdot 3^{n+1}}{\ln(n+1)} \). So, we get:\[\left| \frac{a_{n+1}}{a_n} \right| = \frac{9 \cdot 3^{n+1}}{\ln(n+1)} \times \frac{\ln n}{9 \cdot 3^n} = \frac{3 \ln n}{\ln(n+1)}\]

Step 3: Simplify the Expression

The ratio simplifies to \( \frac{3 \ln n}{\ln(n+1)} \). We consider the limit:\[\lim_{n \to \infty} \frac{3 \ln n}{\ln(n+1)}\]Note that \( \ln(n+1) \approx \ln n + \frac{1}{n} \) for large \( n \). Therefore, it simplifies to:\[\lim_{n \to \infty} \frac{3}{1 + \frac{1}{n \ln n}} = 3 \times \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n \ln n}}\]

Step 4: Evaluate the Limit

As \( n \to \infty \), the term \( \frac{1}{n \ln n} \to 0 \). Hence, the limit becomes:\[\lim_{n \to \infty} \frac{3}{1 + \frac{1}{n \ln n}} = 3 \times 1 = 3\]

Step 5: Determine Convergence or Divergence

The Ratio Test states that if the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \) is less than 1, the series converges absolutely; if \( L > 1 \), the series diverges. Since \( L = 3 \), which is greater than 1, the series \( \sum_{n=2}^{\infty} \frac{3^{n+2}}{\ln n} \) diverges.

Key Concepts

Convergence and Divergence

In mathematics, the concepts of convergence and divergence are crucial when studying infinite series. An infinite series is a sum of infinite terms, and understanding whether it converges or diverges helps mathematicians know if the sum results in a finite value or not.
An infinite series converges if, as we add more and more terms, the total approaches a specific number. If the partial sums of a series (the sums of the first 'n' terms) get closer and closer to a fixed value as 'n' increases, then the series converges to that value.
Conversely, a series diverges if the total does not settle towards a fixed number. This means the series can grow without bound or oscillate without stabilizing.
When conducting series analysis, such as using the Ratio Test, determining convergence or divergence helps in predicting the series' behavior. In our exercise, the series diverges because the limit is greater than 1.

Limit Evaluation

Limit evaluation is the process of finding the value that a function approaches as its input approaches a particular point (often infinity, in the context of series).
In the context of the Ratio Test, we evaluate the limit of the ratio of consecutive terms of a series to determine convergence or divergence. This involves calculating: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
For the given series, evaluating the limit of \( \frac{3 \ln n}{\ln(n+1)} \) as \( n \to \infty \) requires understanding logarithmic approximations for large values. Here, \( \ln(n+1) \approx \ln n + \frac{1}{n} \)
This analysis leads to the simplification of the limit to 3, indicating divergence according to the Ratio Test. Limit evaluation can require manipulation and approximation applying algebraic knowledge to approximate closely.

Infinite Series

An infinite series is a summation of an endless sequence of numbers. Mathematically, this can be expressed as \( \sum_{n=k}^{\infty} a_n \), where \( n \) is the index of summation, starting from some integer \( k \).
Dealing with infinite series requires methods to study their behavior, as adding an infinite number of terms can either lead to a finite value (convergence) or an undefined result (divergence).
In this exercise, the series \( \sum_{n=2}^{\infty} \frac{3^{n+2}}{\ln n} \) is of interest. It is composed of terms with both an exponential component \( 3^n \) and a slowly increasing logarithmic component \( \ln n \).
Understanding series like this involves comparing growth rates and assessing whether the terms shrink fast enough to ensure a finite limit, or grow towards infinity. The Ratio Test assists in this, providing a reliable method to check series for convergence when a straightforward sum is challenging.