Problem 63 - Thomas Calculus

Question

Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(27-90\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\frac{n !}{n^{n}}(\text {Hint} : \text { Compare with } 1 / n .) $$

Step-by-Step Solution

Answer

The sequence converges to 0 as \( n \rightarrow \infty \).

Step 1: Examine the general term

We start by analyzing the given sequence \(a_n = \frac{n!}{n^n}\). To understand its behavior, we need to compare it with a simpler sequence, as suggested in the hint.

Step 2: Simplify the comparison

Consider the sequence \( \frac{1}{n} \) for comparison. Both sequences seem to approach zero as \(n\) grows, thus it's useful to explore if the same behavior holds rigorously for the sequence \( \frac{n!}{n^n} \).

Step 3: Apply Stirling's approximation

Stirling's approximation gives us a way to approximate the factorial: \( n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \). Use this to rewrite the original sequence:\[a_n \approx \frac{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n}{n^n} = \sqrt{2\pi n} \cdot \left(\frac{1}{e}\right)^n.\]

Step 4: Analyze the simplified expression

From the expression derived, \(a_n \approx \sqrt{2\pi n} \cdot \left(\frac{1}{e}\right)^n\). Notice how \(\left(\frac{1}{e}\right)^n\) tends to zero very rapidly, as \(\frac{1}{e} \approx 0.367\), so \(\left(\frac{1}{e}\right)^n\) becomes negligibly small for large \(n\).

Step 5: Conclusion about the convergence

Since \(\left(\frac{1}{e}\right)^n\) diminishes faster than \(\sqrt{2\pi n}\) grows, \(a_n\) approaches zero. Hence, the sequence \(a_n = \frac{n!}{n^n}\) converges to zero.

Key Concepts

Stirling's Approximation

Stirling's approximation is a powerful mathematical tool used to estimate the value of a factorial for large numbers. It provides a simple formula to approximate factorials, especially when direct computation is infeasible due to their large size. This approximation states: \[ n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \]This formula is particularly useful in the analysis of sequences involving factorials. By substituting the factorial in a sequence with its Stirling's approximation, complicated expressions can become more manageable. In our exercise, we used Stirling’s approximation to transform the sequence \( a_n = \frac{n!}{n^n} \) into a simpler form:\[a_n \approx \frac{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n}{n^n} = \sqrt{2\pi n} \cdot \left(\frac{1}{e}\right)^n.\]This simplification helps us analyze the behavior of \( a_n \) as \( n \) approaches infinity. Ultimately, by using Stirling’s approximation, it's easier to determine that the sequence converges to zero.

Factorial Sequence

A factorial sequence involves the factorial operation, which multiplies a series of descending natural numbers. A factorial of a positive integer \( n \) is denoted as \( n! \) and defined as:

• \( n! = n \times (n-1) \times (n-2) \cdots \times 1 \)
• \( 0! \) is defined as \( 1 \)

Factorials grow very quickly as \( n \) increases due to the repeated multiplication. In the context of sequences, such rapid growth can lead to very large numbers, which is why approximations, like Stirling's, are often needed. In our sequence \( a_n = \frac{n!}{n^n} \), the factorial part \( n! \) grows slower than the denominator \( n^n \). This is a crucial observation because the balance between these two rates of growth dictates whether the sequence converges or diverges. By comparing the factorial growth with the power \( n^n \), it becomes evident that the sequence will eventually converge to a specific limit.

Limit of Sequence

The limit of a sequence is the value that the sequence approaches as the index (usually \( n \)) becomes infinitely large. In the analysis of sequences, determining limits is crucial to understanding the long-term behavior of the sequence. For the factorial sequence \( a_n = \frac{n!}{n^n} \), applying Stirling's approximation helped transform and analyze its limit.After substituting Stirling's approximation, the sequence simplifies to:\[ a_n \approx \sqrt{2\pi n} \left(\frac{1}{e}\right)^n \]As \( n \) increases, \( \left(\frac{1}{e}\right)^n \) declines rapidly. This exponential decaying factor diminishes much faster than the potential growth from \( \sqrt{2\pi n} \). Therefore, despite the increasing term \( \sqrt{2\pi n} \), the overall sequence still converges to zero.Studying the limit of sequence shows us not only whether the sequence converges or diverges, but what value or set of values it will approach. For this particular sequence, we confirm that as \( n \to \infty \), the sequence \( a_n \to 0 \). Understanding limits gives insight into the behavior of mathematical models, especially when dealing with factorial sequences or sequences involving large numbers.