Problem 41 - Thomas Calculus

Question

Determining Convergence or Divergence In Exercises \(17-44,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \frac{n ! \ln n}{n(n+2) !}$$

Step-by-Step Solution

Answer

The series converges by the Ratio Test.

Step 1: Write the Given Series

The series given is \( \sum_{n=1}^{\infty} \frac{n! \ln n}{n(n+2)!} \). We need to determine if this series converges or diverges.

Step 2: Simplify the General Term

Simplify the general term \( a_n = \frac{n! \ln n}{n(n+2)!} \). Observe that \( (n+2)! = (n+2)(n+1)n! \), so we can rewrite:\[ a_n = \frac{n! \ln n}{n \cdot (n+2)(n+1)n!} = \frac{\ln n}{(n+2)(n+1)}. \]

Step 3: Analyze the Behavior of the Term

The term \( \frac{\ln n}{(n+2)(n+1)} \) becomes \( \frac{\ln n}{n^2} \) for large \( n \). This is similar to a \( \frac{\ln n}{n^2} \) series, which we should compare with a known convergent series for decision.

Step 4: Apply the Limit Comparison Test

Choose a comparison convergent series, \( \sum \frac{1}{n^2} \), which is known to converge (a p-series with \( p = 2 \)).Calculate the limit:\[ \lim_{n \to \infty} \frac{\frac{\ln n}{n^2}}{\frac{1}{n^2}} = \lim_{n \to \infty} \ln n = \infty. \]Since the limit is infinite, the series does not behave as \( \sum \frac{1}{n^2} \). Try another test for convergence.

Step 5: Apply the Ratio Test

For the ratio test, consider:\[ \lim_{n \to \infty} \left| \frac{\frac{\ln(n+1)}{(n+3)(n+2)}}{\frac{\ln n}{(n+2)(n+1)}} \right| = \lim_{n \to \infty} \frac{\ln(n+1)}{\ln n} \cdot \frac{1}{n+3} \]This simplifies as:\[ \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right) \cdot \frac{1}{n+3} \to 0. \]Since the limit is less than 1, the series converges by the ratio test.

Key Concepts

Limit Comparison Test

The limit comparison test is a powerful tool used to determine whether a series converges or diverges. It works by comparing the series of interest with another series whose convergence properties we already know.

Here's how it generally works:

• Choose a series \( \sum b_n \) that is similar to the given series \( \sum a_n \).
• Calculate the limit: \ \[ \lim_{n \to \infty} \frac{a_n}{b_n} \]
• If this limit \( c \) is a positive, finite number, both series either converge or diverge together.

In our original problem, the series of interest is \( \sum \frac{\ln n}{n^2} \). We compared it with the convergent p-series \( \sum \frac{1}{n^2} \), which has \( p = 2 \). However, the limit resulted in infinity, indicating that the series \( \sum \frac{\ln n}{n^2} \) diverges based on this test.

This led us to look for another method to check for convergence, as this test was inconclusive.

Ratio Test

The ratio test is another method to determine the convergence or divergence of a series. It focuses on the relative sizes of consecutive terms in the series.

To apply the ratio test, follow these steps:

• Compute the ratio of the absolute values of consecutive terms: \( \left| \frac{a_{n+1}}{a_n} \right| \).
• Take the limit as \( n \) approaches infinity: \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]
• Determine the outcome:
  • If \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \) or \( L \) is infinite, the series diverges.
  • If \( L = 1 \), the test is inconclusive.

In our specific series, this method worked well. After simplifying, the limit was zero, which means the series converges.

The ratio test successfully confirmed the convergence of our series, where other methods weren't sufficient.

Factorial Simplification

The simplification of factorials is often a key step when dealing with series that involve factorial expressions. Simplifying factorial terms can make convergence tests, like the ratio and limit comparison tests, more manageable.

Let's look at a basic approach:

• Factorials grow very rapidly, and it helps to break them into smaller, more manageable parts.
• In our series, the term \( (n+2)! \) simplifies as \( (n+2)(n+1)n! \).
• This simplification allowed us to cancel \( n! \) from the numerator and denominator, simplifying further steps.

By reducing the factorial to a polynomial-like structure, you can apply standard series convergence tests more easily.

This simplification was crucial in finding convergent behavior or in comparing the series with more familiar forms.