Problem 38 - Thomas Calculus

Question

In Exercises \(17-46,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$ \sum_{n=1}^{\infty} \frac{n !}{(-n)^{n}} $$

Step-by-Step Solution

Answer

The series diverges by the Ratio Test as the terms grow rapidly.

Step 1: Understanding the Series

We are given the series \( \sum_{n=1}^{\infty} \frac{n !}{(-n)^{n}} \). The task is to determine if this infinite series converges or diverges. Analyzing the terms, the numerator is \( n! \) (n factorial) and the denominator is \((-n)^n\), which means the terms are \( \frac{n!}{(-n)^n} \). Due to the factorial in the numerator, the terms grow fast.

Step 2: Analyzing the Behavior of Terms

Examine the general term \( a_n = \frac{n!}{(-n)^n} \). To determine convergence or divergence, observe its behavior as \( n \) approaches infinity. The factorial \( n! \) grows extremely fast, while \( (-n)^n \) is a large power (with an alternating negative sign depending on \( n \)). However, the term's magnitude is more influenced by \( n! \), which usually grows faster than polynomial terms like \((-n)^n\).

Step 3: Ratio Test Application

Use the Ratio Test, which is suitable when dealing with factorials. Compute the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). We have:\[ a_{n+1} = \frac{(n+1)!}{(-(n+1))^{n+1}} \]With\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(-(n+1))^{n+1}} \times \frac{(-n)^n}{n!} = \frac{(n+1)\times n!}{(-(n+1))^{n+1}} \times \frac{(-n)^n}{n!} \]Simplifying, this becomes:\[ \frac{n+1}{-((n+1)/n)^n \times (n+1)} = \frac{1}{(-1)^{n+1}} \times \left(\frac{n}{n+1}\right)^n \approx \frac{1}{-1} \]This calculation shows that this ratio is not tending to zero or having properties of convergence.

Step 4: Conclusion from Ratio Test

In the Ratio Test, the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \) is not a value less than 1. In fact, the factorial growth outpaces the polynomial in the denominator, suggesting divergence of the series. Since the terms do not approach zero and the ratio test limit behaves unfavorably for convergence, we conclude that the series diverges.

Key Concepts

Ratio Test

The ratio test is an efficient method to determine whether a series converges or diverges. It is particularly helpful when dealing with series involving factorials or exponential terms, as it simplifies complex expressions.
To apply the ratio test:

• Consider the general term of the series: \( a_n \).
• Compute the limit \( L = \lim_{{n \to \infty}} \left| \frac{{a_{n+1}}}{a_n} \right| \).
• Analyze the value of \( L \):
• If \( L < 1 \), the series converges.
• If \( L > 1 \) or \( L \) is infinite, the series diverges.
• If \( L = 1 \), the test is inconclusive.

For our case, the Limit \( L \) did not tend to a value less than 1, indicating that the series diverges. The factorial in the numerator grows rapidly, making the series diverge.

Factorial Function

The factorial function, denoted as \( n! \), is the product of all positive integers up to \( n \). It grows very quickly, especially for larger \( n \).
For example:

• \( 3! = 3 \times 2 \times 1 = 6 \)
• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

Factorials grow faster than exponential and polynomial expressions, which makes them interesting in convergence tests.
In our series, the factorial \( n! \) growth is crucial. When competing with a polynomial-like expression in the denominator, it eventually dominates, leading to divergence.

Power Series

A power series is an infinite series of the form \( \sum_{n=0}^{\infty} a_n x^n \). This consists of terms involving a variable raised to an increasing power and multiplied by coefficients.

• A classic example is the geometric series \( \sum_{n=0}^{\infty} x^n \).
• Power series represent functions and appear in calculus and analysis, providing approximations to functions such as \( e^x \) or \( \sin(x) \).

In convergence tests, power series are examined to see whether they sum to a specific value or diverge.
In our exercise, though not a simple power series, the term \( (-n)^n \) plays a role similar to a power series, influencing the overall behavior significantly.

Infinite Series

An infinite series is the sum of infinitely many terms: \( \sum_{n=1}^{\infty} a_n \). The core idea is to see if this sum approaches a limit, i.e., converges, or does not settle, i.e., diverges.
Infinite series are fundamental in mathematical analysis, modeling, and applications. The challenge is to evaluate whether the accumulated sum nears a finite value or keeps growing.
To determine convergence or divergence, various tests like the ratio test, comparison test, and integral test, are used.
Focusing on our series \( \sum_{n=1}^{\infty} \frac{n!}{(-n)^n} \), the rapid growth of \( n! \) versus the alternating power term in the denominator decides its divergent nature. While it might seem manageable, factorial growth is enormous.