Problem 42 - Single Variable Calculus

Question

Prove that \(\lim _{n \rightarrow \infty} \frac{2^{n} n !}{n^{n}}=0\). Hint: Show that \(\sum_{n=1}^{\infty} \frac{2^{n} n !}{n^{n}}\) is convergent.

Step-by-Step Solution

Answer

To prove that \(\lim _{n \rightarrow \infty} \frac{2^{n} n !}{n^{n}}=0\), we first apply the Root Test to the series \(\sum_{n=1}^{\infty} \frac{2^{n} n !}{n^{n}}\), and find the limit of the nth root of the terms. Applying L'Hôpital's Rule and noting that \(n!\) grows much faster than \((n!)^{1/n}\), we find that the limit is 0. Since this limit is less than 1, the series is convergent, and therefore the limit of its terms is 0. This concludes our proof.

Step 1: Apply the Root Test to the series

We'll apply the Root Test to the series \(\sum_{n=1}^{\infty} \frac{2^{n} n !}{n^{n}}\). To do this, consider the absolute value of the terms and find the limit of the nth root: \[\lim_{n\to\infty} \left|\frac{ 2^n n!}{n^n} \right|^{1/n} = \lim_{n\to\infty} \frac{2^{n/n} (n!)^{1/n}}{n} \]

Step 2: Calculate the limit of the nth root of the terms

Now, we calculate the limit: \[ \lim_{n\to\infty} \frac{2^{n/n} (n!)^{1/n}}{n} = \lim_{n\to\infty} \frac{2 (n!)^{1/n}}{n} \] Applying L'Hôpital's Rule for the last limit, we have: \[ \lim_{n\to\infty} \frac{2 (n!)^{1/n}}{n} = \lim_{n\to\infty} \frac{2 \frac{1}{n} (n!)^{1/n - 1} n!}{1} = 2 \lim_{n\to\infty} (n!)^{1/n - 1} \] Since \(n!\) grows much faster than \((n!)^{1/n}\) it implies \((n!)^{1/n - 1}\) is getting smaller as \(n\) grows. Therefore the limit is: \[ 2 \lim_{n\to\infty} (n!)^{1/n - 1} = 0 \] Now, by the Root Test, because this limit is less than 1, the series is convergent.

Step 3: Conclude the limit of the terms of the series

Since the series \(\sum_{n=1}^{\infty} \frac{2^{n} n !}{n^{n}}\) is convergent, that means the limit of its terms is 0: \[\lim _{n \rightarrow \infty} \frac{2^{n} n !}{n^{n}}=0\] This concludes our proof.

Key Concepts

Convergence of Series

When we talk about the convergence of a series, we are essentially exploring whether the sum of an infinite series leads to a finite number. This happens when the sum of the terms becomes closer and closer to a specific value as we keep adding more terms. Convergence is crucial in calculus and real analysis, as it lets us know if an infinite sum makes sense numerically.
For the series \( \sum_{n=1}^{\infty} \frac{2^{n} n !}{n^{n}}\), convergence is determined by testing the limit of its terms. If the limit approaches zero, the series is typically convergent.

• In our example, showing that this series converges helps us understand that the limit of its terms will indeed approach zero, validating the original limit question.

Understanding convergence involves familiar tools, such as limits and tests like the Root Test, which we use to confirm the behavior of our series as \(n\) approaches infinity.

Factorials

Factorials are a mathematical function found in many areas of mathematics, especially in combinatorics, algebra, and calculus. The factorial of a non-negative integer \(n\) is the product of all positive integers less than or equal to \(n\). It is denoted by \(n!\). For instance, \(3! = 3 \times 2 \times 1 = 6\).
Factorials grow very rapidly as \(n\) increases. For example, \(5! = 120\) and by \(10!\), the value jumps to 3,628,800! This explosive growth plays a key role in limit problems and series analysis.

• In our exercise, \(n!\) is part of the term \(\frac{2^n n!}{n^n}\). The rapid growth of \(n!\), compared to the denominator \(n^n\), helps us see that the limit approaches zero.

The concept of factorials is fundamental for understanding why certain series converge, especially when combined with powers and exponential terms.

Root Test

The Root Test, also known as the Cauchy Root Test, is a powerful tool for analyzing the convergence of series. It considers the \(n\)-th root of the terms in a series and evaluates their behavior as \(n\) tends to infinity. It is specifically useful when dealing with series involving factorials and exponential terms.

• For a series \(\sum a_n\), the Root Test looks at \(\lim_{n \to \infty} \sqrt[n]{|a_n|}\).
• If this limit \(< 1\), the series is absolutely convergent.
• If the limit \(> 1\), the series diverges.
• If the limit equals \(1\), the test is inconclusive.

In our problem, applying the Root Test to \(\sum_{n=1}^{\infty} \frac{2^{n} n !}{n^{n}}\) helped confirm that the limit of \(\frac{2^n n!}{n^n}\) is zero, as the terms diminish faster with the \(n\)-th root considered, ultimately leading us to conclude convergence.