Question

Determine the convergence of the given series. State the test used; more than one test may be appropriate. $$\sum_{n=1}^{\infty} \frac{n ! n ! n !}{(3 n) !}$$

Short answer

The series converges by the Ratio Test.

Step-by-step solution

Identify the Series

The series given is \( \sum_{n=1}^{\infty} \frac{n! n! n!}{(3n)!} \). We want to determine whether this series converges and, if so, find a suitable test to do so.

Apply the Ratio Test

The Ratio Test is suitable here. For the sequence \( a_n = \frac{n! n! n!}{(3n)!} \), compute the ratio \( \left| \frac{a_{n+1}}{a_n} \right| \).\[\frac{a_{n+1}}{a_n} = \frac{((n+1)!)^3}{(3(n+1))!} \cdot \frac{(3n)!}{(n!)^3} = \frac{(n+1)^3}{(3n+1)(3n+2)(3n+3)}\]

Simplify the Ratio

Simplify \(\frac{(n+1)^3}{(3n+1)(3n+2)(3n+3)}\) by expanding and analyzing its behavior as \( n \to \infty \). The cubic terms grow as:\[\frac{(n+1)^3}{27n^3} = \frac{1}{27} + \text{lower order terms} \] As \( n \to \infty \), the lower order terms become negligible.

Evaluate the Limit

Find \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). With the simplified expression:\[L = \lim_{n \to \infty} \frac{1}{27} + \text{negligible terms} = \frac{1}{27}\]Since \( L < 1 \), the Ratio Test indicates that the series converges.

Key concepts

Ratio Test

The Ratio Test is a powerful tool used to determine the convergence of infinite series. Let's delve into how it works and why it's relevant. For a given series with terms \(a_n\), the test involves examining the limit of the absolute value of the ratio of consecutive terms:

• The series \(\sum a_n\) converges if \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L < 1\).
• If \(L > 1\) or the limit does not exist, the series diverges.
• If \(L = 1\), the test is inconclusive.

In our exercise, we calculated:

1. \(\left| \frac{a_{n+1}}{a_n} \right|\) using \(a_n = \frac{n! n! n!}{(3n)!}\).
2. The expression simplified to \(\frac{(n+1)^3}{(3n+1)(3n+2)(3n+3)}\).

By simplifying further, we found \(L = \frac{1}{27}\), which is less than 1, indicating the series converges according to the Ratio Test.

Factorials in Series

Factorials, indicated by the symbol \(!\), play a crucial role in series. They represent products of all positive integers up to a given number \(n! = n \times (n-1) \times \ldots \times 1\). Using factorials in series often results in rapidly growing terms.
When analyzing series like \(\sum \frac{n! n! n!}{(3n)!}\), factorials determine both the growth and complexity of the sequence.

• The product \(n! n! n!\) in the numerator grows very quickly with each increasing term of the series.
• The denominator \((3n)!\) grows even faster since it involves more factors (i.e., up to \(3n\)).

This contrast in growth rates between the numerator and denominator is crucial for convergence analysis. When used with tests like the Ratio Test, the behavior of factorials helps us determine whether the terms decrease sufficiently quickly enough for the entire series to converge.

Limits and Continuity

In the context of series, limits help us understand the behavior of terms as \(n\) approaches infinity. Specifically, when employing tests like the Ratio Test, we're often interested in finding the limit of the ratio of consecutive terms.
For example, in our series, the goal was to find \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\). This requires analyzing the simplified formula:

• Beyond comparing terms, limits allow us to assess whether terms grow or shrink endlessly.
• Once the limit is understood, it gives a conclusive answer regarding convergence.

Concepts from calculus, like limits and continuity, underline the predictability and behavior of sequences. Such analysis is pivotal in reaching rigorous conclusions about series convergence and the nature of their terms.