Problem 62 - Thomas Calculus

Question

In Exercises \(57 - 82 ,\) use any method to determine whether the series converges or diverges. Give reasons for your answer. $$ \sum _ { n = 2 } ^ { \infty } \frac { ( 3 n ) ! } { ( n ! ) ^ { 3 } } $$

Step-by-Step Solution

Answer

The series \( \sum_{n=2}^{\infty} \frac{(3n)!}{(n!)^3} \) diverges by the Ratio Test.

Step 1: Understand the Series

We are given the series \( \sum_{n=2}^{\infty} \frac{(3n)!}{(n!)^3} \). This is a hypergeometric-like series, and we need to determine its convergence or divergence.

Step 2: Apply the Ratio Test

The Ratio Test involves finding the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \), where \( a_n = \frac{(3n)!}{(n!)^3} \). If \( L < 1 \), the series converges; if \( L > 1 \), it diverges; if \( L = 1 \), the test is inconclusive.

Step 3: Express Terms for the Ratio Test

Compute \( a_{n+1} = \frac{(3(n+1))!}{((n+1)!)^3} \). Substitute these into the ratio \( \frac{a_{n+1}}{a_n} \) to get \( \frac{(3(n+1))!}{((n+1)!)^3} \cdot \frac{(n!)^3}{(3n)!} \).

Step 4: Simplify the Ratio

Expand \((3(n+1))! = (3n+3)(3n+2)(3n+1)(3n)!\) and \((n+1)! = (n+1) \cdot n!\). Substitute these back into \(\frac{(3(n+1))! / (n!)^3}{((n+1)!)^3 / (n!)^3}\) to simplify the expression.

Step 5: Compute the Ratio Limit

The ratio simplifies to: \[ \lim_{n \to \infty} \frac{(3n+3)(3n+2)(3n+1)}{(n+1)^3} \]. Evaluate this limit as \( n \rightarrow \infty \). The leading terms \( (3n)^3 \) dominate, giving \[ \lim_{n \to \infty} \frac{27n^3}{n^3} = 27 \].

Step 6: Conclusion on Convergence

Since the limit \( L = 27 \) (which is greater than 1), the Ratio Test tells us that the series \( \sum_{n=2}^{\infty} \frac{(3n)!}{(n!)^3} \) diverges.

Key Concepts

Understanding the Ratio Test

The Ratio Test is a popular tool used to determine the convergence of series. It is particularly useful when dealing with series where each term involves factorials, exponentials, or powers. In essence, the test involves taking the ratio of the \(n+1\)th term to the \(n\)th term of a series and then finding the limit as \(n\) approaches infinity. This limit is denoted as \(L\). The steps involved in applying the Ratio Test are:

• Determine the \(n\)th term, \((a_n)\), of the series.
• Compute the \(n+1\)th term, \(a_{n+1}\).
• Find the absolute value of the ratio \(|a_{n+1}/a_n|\).
• Evaluate the limit \(L = \lim_{{n \to \infty}} |a_{n+1}/a_n|\).

If \(L < 1\), the series converges. If \(L > 1\), it diverges. When \(L = 1\), the test is inconclusive, and another convergence test may be needed.

Understanding Series Divergence

In mathematics, series divergence indicates that a series does not have a finite sum. Rather than settling to a specific value, the sum of its terms grows indefinitely. Determining whether a series diverges or converges is essential, particularly in calculus, where infinite series play a significant role. Divergence often occurs in series with terms that do not decrease rapidly enough.

The Ratio Test, as applied in the original problem, indicated divergence because the limit of the series' term ratio was greater than 1, specifically 27. This highlights that the terms do not decrease sufficiently fast, causing the series to diverge. Divergent series can appear similar to convergent ones initially, but their terms eventually accumulate without bound.

Working with Factorial Series

Factorial series involve series with terms expressed in terms of factorials. These series often present complexity due to the rapid growth of factorials, as \(n! = n \times (n-1) \times (n-2) \times \ldots \times 1\). The terms grow much faster than polynomial or exponential terms, particularly as \(n\) increases.

In the specific exercise, the presence of the factorial \( (3n)! \) in the numerator and \( (n!)^3 \) in the denominator dynamically affects the growth rate of each term, with the series quickly increasing as \(n\) becomes larger. Factorial series often necessitate careful manipulation, such as the Ratio Test, to understand their behavior. Luckily, the Ratio Test is well-suited for factorial series, as these allow simplifications that make it easier to determine convergence or divergence.