Problem 41 - Thomas Calculus

Question

In Exercises \(17-46,\) 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 direct comparison test.

Step 1: Analyze the Series Expression

The given series is \( \sum_{n=1}^{\infty} \frac{n ! \ln n}{n(n+2) !} \). Notice the presence of \(n!\) in the numerator and \((n+2)!\) in the denominator. This suggests that the terms decrease quickly as they are dominated by the factorial in the denominator.

Step 2: Simplify the Terms

Simplify the expression \( a_n = \frac{n! \ln n}{n(n+2)!} \) by canceling \(n!\) with part of \((n+2)!\). Thus, \( a_n = \frac{\ln n}{(n+2)(n+1)} \).

Step 3: Apply Comparison Test

Consider \( b_n = \frac{\ln n}{n^3} \) for comparison, assuming \(n > 2\). Compare \( a_n \) and \( b_n \): \( \frac{\ln n}{(n+2)(n+1)} < \frac{\ln n}{n^3} \) for large \(n\). Since the series \( \sum \frac{\ln n}{n^3} \) converges (by integral test or comparison to \(\sum \frac{1}{n^2} \)), \( \sum a_n \) also converges by the direct comparison test.

Step 4: Conclusion

By the direct comparison test, since \( a_n < b_n \) and \( \sum b_n \) converges, \( \sum a_n = \sum_{n=1}^{\infty} \frac{n ! \ln n}{n(n+2) !} \) also converges.

Key Concepts

Factorial Notation in Series

Factorial notation often appears in series, representing the product of all positive integers up to a specified number. It is written as \( n! \), which means \( 1 \times 2 \times 3 \times \ldots \times n \). Factorial terms grow very quickly because with each additional number in the sequence, the product becomes much larger.
The given series \( \sum_{n=1}^{\infty} \frac{n! \ln n}{n(n+2)!} \) incorporates factorial notation both in the numerator and the denominator. Despite the numerically larger factorial \( n! \) in the numerator, \( (n+2)! \) at the denominator means that the factorial in the denominator grows even faster because it continues past \( n \), growing with even more factors. This results in the terms of the series decreasing rapidly, influencing the convergence behavior of the series.

Comparison Test

The comparison test is a useful method when determining the convergence of a series. It involves comparing the series of interest to a second, more understood series. Here's how it works:

• If a series \( \sum a_n \) is less than or equal to a convergent series \( \sum b_n \) term by term for all \( n \), then \( \sum a_n \) converges.
• Conversely, if a series \( \sum a_n \) is greater than or equal to a divergent series \( \sum b_n \) term by term, \( \sum a_n \) diverges.

The series in question is \( \sum_{n=1}^{\infty} \frac{\ln n}{(n+2)(n+1)} \). We compared this to the series \( \sum_{n=1}^{\infty} \frac{\ln n}{n^3} \), a known convergent series using integration techniques.
By showing that each term in our original series is less than or equal to the corresponding term in the convergent series, the comparison test tells us that the series converges.

Integral Test

The integral test can be applied to determine if a series converges. It is based on integrating a function that resembles the terms of the series and looking at the behavior of this integral.
For a series \( \sum_{n=1}^{\infty} a_n \) where \( a_n = f(n) \) corresponds closely to a continuous, positive, decreasing function \( f(x) \) for \( x > N \), we use the test:

• If the improper integral \( \int_{N}^{\infty} f(x)\,dx \) converges, then \( \sum a_n \) also converges.
• If the integral diverges, so does \( \sum a_n \).

In our comparison to \( \sum \frac{\ln n}{n^3} \), we rely on the fact that similar integrals, such as \( \int_{N}^{\infty} \frac{\ln x}{x^3}\,dx \), converge, providing us further assurance of the behavior of the series.

Series Simplification

Simplifying series expressions is often necessary to make sense of their convergence. Originally, the series was given as \( \sum_{n=1}^{\infty} \frac{n! \ln n}{n(n+2)!} \).
To simplify it, we start by analyzing the role of factorials. The expression \( n! \) from the numerator and a part of \( (n+2)! \) from the denominator can be canceled out:

\[ a_n = \frac{n! \ln n}{n(n+2)!} = \frac{\ln n}{(n+2)(n+1)} \]

This simplification aids in applying tests such as the comparison test and helps one realize that factorial growth in the denominator will outstrip that of the numerator, resulting in smaller terms. Such simplification is crucial because it allows the series to be more easily compared to standard series known for their convergence behavior.