Problem 32 - Calculus: Early Transcendentals

Question

Test the series for convergence or divergence. \( \displaystyle \sum_{n = 1}^{\infty} \frac {(n!)^n}{n^{4n}} \)

Step-by-Step Solution

Answer

The series converges.

Step 1: Recognize the Series Format

The series in question is \( \sum_{n = 1}^{\infty} \frac{(n!)^n}{n^{4n}} \). This series appears to have factorials and powers involved, making it a prime candidate for the Ratio Test, which works well with factorials.

Step 2: Apply the Ratio Test

The Ratio Test involves examining the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \), where \( a_n = \frac{(n!)^n}{n^{4n}} \). This test will help us determine if the series converges absolutely, diverges, or if the test is inconclusive.

Step 3: Compute the Ratio of Successive Terms

Calculate \( \frac{a_{n+1}}{a_n} \):\[\frac{a_{n+1}}{a_n} = \frac{\left((n+1)!\right)^{n+1}}{(n+1)^{4(n+1)}} \cdot \frac{n^{4n}}{(n!)^n} = \frac{((n+1)!)^{n+1}}{n!^n} \cdot \frac{n^{4n}}{(n+1)^{4(n+1)}}.\]

Step 4: Simplify the Ratio

Simplify the expression:The factorial portion expands to:\[ \frac{((n+1)!)^{n+1}}{(n!)^n} = ((n+1)!)^{n+1-n} = (n+1) \cdot (n+1)!\]However, note that:\[ \frac{((n+1)!)^{n+1}}{(n!)^n} = \left(\frac{(n+1)!}{n!}\right)^n (n+1) = (n+1)^n (n+1) = (n+1)^{n+1}.\]Substituting back, we have:\[ \frac{(n+1)^{n+1}}{(n+1)^{4(n+1)}} \cdot n^{4n} = \frac{(n+1)^{n+1} \cdot n^{4n}}{(n+1)^{4n+4}}.\]

Step 5: Simplify Further

Expand and collect terms:\[ = \frac{(n+1)^{n+1}}{(n+1)^{4n+4}} \cdot \left(\frac{n}{n+1}\right)^{4n} = \frac{1}{(n+1)^{4(n+1) - (n+1)}} \cdot \left(\frac{n}{n+1}\right)^{4n}.\]Thus:\[ \frac{1}{(n+1)^{3n+3}} \cdot \left(\frac{n}{n+1}\right)^{4n} = \frac{1}{(n+1)^{3n+3}} \cdot \left( \frac{n}{n+1} \right)^{4n}.\]

Step 6: Determine the Limit as n Approaches Infinity

As \( n \to \infty \), the fraction \( \left(\frac{n}{n+1}\right)^{4n} \to e^{-4} \) due to the limit of form \( \left(1 - \frac{1}{n+1}\right)^{n+1} = \frac{1}{e}.\) Hence \( \left(\frac{n}{n+1}\right)^{4n} \) tends to \( e^{-4} \).This vanishes when combined with \( \frac{1}{(n+1)^{3n+3}} \), leading \( \lim_{n \to \infty} \frac{1}{(n+1)^{3n+3}} \cdot e^{-4} = 0.\)

Step 7: Conclude with the Ratio Test

Because the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 \), which is less than 1, the series \( \sum_{n=1}^{\infty} \frac{(n!)^n}{n^{4n}} \) converges by the Ratio Test.

Key Concepts

Factorials in Series

Factorials are an essential component in many mathematical problems, particularly with series. In a series, factorials often present as product terms such as \( n! \), which could be read as "n factorial." Essentially, this represents the product of all positive integers from 1 to \( n \).
Factorials tend to grow very rapidly compared to other terms in a series, especially when \( n \) becomes large. For example, even for small values like \( n = 5 \), \( 5! = 120 \), which is quite large considering \( n \) itself is only 5. This extraordinarily fast growth can influence the convergence of an infinite series greatly.

• Expressions involving factorials, like \( (n!)^n \), grow even faster since it's raising a factorial to a power.
• This growth rate is crucial in determining if an infinite series converges or diverges.

Understanding how factorials behave in such series is important for applying convergence tests correctly.

Ratio Test

The Ratio Test is a popular method for determining the convergence of an infinite series. It is particularly handy when dealing with series that involve factorials or exponentiated terms.
The test considers the limit:
\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]where \( a_n \) is the general term of the series.

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \), the series diverges.
• If \( L = 1 \), the test is inconclusive.

For the given series, the term \( a_n = \frac{(n!)^n}{n^{4n}} \), involves complex factorials and power terms.
After simplifications, finding that the limit \( L \) ended up as 0, which is less than 1, helped decisively categorize the series as converging.

Infinite Series Convergence

The convergence of an infinite series deals with whether the sum of its infinite terms approaches a finite number. This concept is important in mathematics because it tells us if adding more terms will stabilize to a set value or not.

• An infinite series is written as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) are the individual terms.
• The goal is to assess whether its total sum becomes fixed (convergence) or not (divergence).

Several methods exist for checking convergence, such as the Ratio Test, the Root Test, or the Integral Test.
Analyzing the nature of the terms \( a_n \), especially when involving factorials or powers, is crucial for choosing the right test.
In this exercise, understanding the rapid growth caused by factorials and applying the right convergence tests resulted in determining the outcome of the series as convergent.