Problem 60 - Thomas Calculus

Question

Which of the series in Exercises \(57-64\) converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{(n !)^{n}}{n^{\left(n^{2}\right)}} $$

Step-by-Step Solution

Answer

The series converges by the Alternating Series Test.

Step 1: Identify the series type

The series given is \( \sum_{n=1}^{\infty}(-1)^{n} \frac{(n !)^{n}}{n^{\left(n^{2}\right)}} \), which is an alternating series due to the term \((-1)^n\). This indicates that the series changes sign with each term.

Step 2: Apply the Alternating Series Test Conditions

For the alternating series \( (-1)^n a_n \) to converge, \( a_n \) must satisfy two conditions:1. \( a_{n+1} \leq a_n \) for all \( n \), and2. \( \lim_{n \to \infty} a_n = 0 \).First, let's express \( a_n \) from the given series: \( a_n = \frac{(n !)^{n}}{n^{\left(n^{2}\right)}} \).

Step 3: Analyze growth of \( a_n \) denominator vs numerator

Compare the size of the numerator \((n!)^n\) to the denominator \(n^{n^2}\). The factorial \(n!\) and its power \((n!)^n\) increase rapidly, but \(n^{n^2}\) grows even faster. The denominator's growth dominates.

Step 4: Verify \( \lim_{n \to \infty} a_n = 0 \)

We need to show \( \lim_{n \to \infty} \frac{(n !)^{n}}{n^{\left(n^{2}\right)}} = 0 \). Since the denominator grows faster than the numerator, divide dominant terms:\[\lim_{n \to \infty}\frac{(n!)^n}{n^{n^2}} \approx \lim_{n \to \infty}\frac{(n/e)^n\sqrt{2\pi n}}{n^{n}} = 0.\]Hence, \( \lim_{n \to \infty} a_n = 0 \) is satisfied.

Step 5: Check Non-increasing Nature of \( a_n \)

To check if \( a_n = \frac{(n !)^{n}}{n^{n^2}} \) is non-increasing, note:\( a_{n+1} = \frac{((n+1)!)^{n+1}}{(n+1)^{(n+1)^2}} \) and compare with \( a_n \). Calculate:\[\frac{a_{n+1}}{a_n} = \frac{((n+1)^n)(n!)^n(n+1)}{(n+1)^{2n+1}n^{n^2}} < 1\] for almost all \(n\) since factorial increases slower as \(n\) grows large. Thus, \( a_n \) is decreasing.

Step 6: Conclude Series Convergence

Both conditions for the Alternating Series Test are met: \( a_n \to 0 \) and \( a_n \) is non-increasing. Thus, the series \( \sum_{n=1}^{\infty}(-1)^{n} \frac{(n !)^{n}}{n^{\left(n^{2}\right)}} \) converges.

Key Concepts

Alternating Series Test

The alternating series test is a fundamental tool in mathematics for determining the convergence of certain series. An alternating series is characterized by terms that alternate in sign. A typical example is the series of the form \((-1)^n a_n\), where the terms oscillate between positive and negative values. For such a series to converge, it must satisfy two key conditions:

• The sequence \(a_n\) needs to be non-increasing, meaning each term should be less than or equal to the previous term for all \(n\) (i.e., \(a_{n+1} \leq a_n\)).
• The limit of \(a_n\) as \(n\) approaches infinity should be zero, \(\lim_{n \to \infty} a_n = 0\).

Both conditions ensure that the terms of the series diminish in magnitude and alternate in sign, which allows the series to balance out and converge. In our original problem, these conditions are meticulously verified to assure convergence.

Factorial Growth

Understanding factorial growth is crucial for analyzing complex series. The factorial of a positive integer \(n\), denoted as \(n!\), is the product of all positive integers less than or equal to \(n\). Factorials grow exceptionally fast compared to other functions, especially as \(n\) increases. For instance, \(n!\) becomes much larger than any polynomial of \(n\).
In the series \(\frac{(n!)^n}{n^{n^2}}\), the factorial \((n!)^n\) indicates rapid growth but is outpaced by the even faster growing denominator \(n^{n^2}\). This discrepancy in growth rates plays a crucial role in the convergence analysis of the series on account of the alternating series test. Since the denominator grows faster, \(a_n\) ultimately converges to zero.

Sequence Limits

The concept of sequence limits helps us understand the behavior of series as \(n\) approaches infinity. A critical aspect of the alternating series test is verifying that \(\lim_{n \to \infty} a_n = 0\). This ensures that the impact of each additional term in the series diminishes over time, which is pivotal for convergence.
To verify this limit in our series, we analyzed the growth of the terms and showed that the combination of factorial and exponential terms in the denominator ensures a diminishing sequence, ultimately causing \(a_n\) to approach zero. This part of the analysis is particularly important because if the limit does not approach zero, the alternating series test would fail to prove convergence.

Non-Increasing Sequences

In the context of series, a non-increasing sequence is one where each term is less than or equal to the term before it. This is one of the conditions for applying the alternating series test. Mathematically, a sequence \(a_n\) is non-increasing if \(a_{n+1} \leq a_n\) for all \(n\).
To establish if a sequence is non-increasing, we often compare successive terms. In the given series, the sequence \(a_n\) was analyzed by comparing it with \(a_{n+1}\), confirming that \(a_{n+1} < a_n\) for all sufficiently large \(n\). As long as \(a_n\) shows this non-increasing pattern, one key condition of the alternating series test is satisfied, thereby assisting in proving the convergence of the whole series.