Problem 73 - 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. $$ a_{n}=\frac{3^{n} \cdot 6^{n}}{2^{-n} \cdot n !} $$

Step-by-Step Solution

Answer

The sequence converges to 0.

Step 1: Simplify the Sequence

First, simplify the given sequence \(a_{n}\). The given expression is \(a_{n} = \frac{3^{n} \cdot 6^{n}}{2^{-n} \cdot n!}\). Notice that \(6^{n} = (2 \cdot 3)^{n} = 2^{n} \cdot 3^{n}\). Substitute this into the expression:\[ a_{n} = \frac{3^{n} \cdot (2^{n} \cdot 3^{n})}{2^{-n} \cdot n!} = \frac{3^{2n} \cdot 2^{n}}{2^{-n} \cdot n!} \]Thus, the expression becomes:\[ a_{n} = \frac{3^{2n} \cdot 2^{n}}{2^{-n} \cdot n!} = 3^{2n} \cdot \frac{2^{n}}{2^{-n}} \cdot \frac{1}{n!} \]Notice that \(\frac{2^{n}}{2^{-n}} = 2^{2n} \), so the expression further simplifies to:\[ a_{n} = \frac{3^{2n} \cdot 2^{2n}}{n!} = \frac{(6^2)^n}{n!} = \frac{36^n}{n!} \].

Step 2: Apply the Ratio Test

To determine the convergence or divergence of the sequence, we apply the Ratio Test:Calculate \(\lim\limits_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right|\). The terms of the sequence are \(a_{n} = \frac{36^n}{n!}\) and \(a_{n+1} = \frac{36^{n+1}}{(n+1)!}\).Ratio of consecutive terms:\[\frac{a_{n+1}}{a_{n}} = \frac{36^{n+1}}{(n+1)!} \cdot \frac{n!}{36^n} = \frac{36 \cdot 36^n}{36^n} \cdot \frac{n!}{(n+1) \cdot n!} = \frac{36}{n+1}\].Take the limit as \(n\) approaches infinity:\[\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right| = \lim_{n \to \infty} \frac{36}{n+1} = 0 \].

Step 3: Conclusion from the Ratio Test

Since \(\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right| = 0\) and this limit is less than 1, the Ratio Test tells us that the sequence \(\left\{a_n\right\}\) converges.

Step 4: Find the Limit of the Sequence

The sequence \(a_n = \frac{36^n}{n!}\) converges to 0. This is evident because as \(n\) becomes very large, the factorial \(n!\) in the denominator grows much faster than the exponential term \(36^n\) in the numerator, driving the fraction towards 0.

Key Concepts

Convergence and Divergence

When we talk about the limit of a sequence, we focus on whether the terms of the sequence settle down to a particular number, or if they continue to vary. This is what we consider when determining convergence and divergence.

• **Convergent sequence:** If the terms of the sequence approach a specific limit as the index number grows indefinitely, the sequence is convergent. For instance, if you have a series of fractions and as they progress, they consistently get closer to a number, say 1, you can say the sequence converges to 1.
• **Divergent sequence:** A sequence that doesn't approach a specific limit behaves erratically as the index increases or keeps increasing without bounds is divergent. An example of a divergent sequence would be one that grows larger without settling towards any particular value.

Understanding whether a sequence converges or diverges helps in studying its behavior in mathematical analysis, such as the case with the series given by \( a_{n} = \frac{36^n}{n!} \). Here, convergence means the terms are eventually drawn toward 0. This happens because the factorial in the denominator increases much faster than the exponential term in the numerator.

Ratio Test

The Ratio Test is a handy tool in mathematics to determine if an infinite series converges or diverges. It's especially useful when dealing with sequences involving factorial terms or powers, such as in the formula \( a_{n} = \frac{36^n}{n!} \). Here's how you use it:

• **Step 1:** Form the ratio \( \frac{a_{n+1}}{a_{n}} \). This represents the ratio of the subsequent term to the current term.
• **Step 2:** Calculate the absolute limit: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}} \right| \).

If the limit is less than 1, the series converges. If it's more than 1, the series diverges. If it equals 1, the test is inconclusive.
In the given sequence, the ratio \( \frac{a_{n+1}}{a_{n}} \) simplifies to \( \frac{36}{n+1} \), and the limit as \( n \) approaches infinity is 0, which is clearly less than 1. Thus, the sequence converges. This test is crucial because it gives a straightforward criterion to determine convergence for sequences which might seem complex to evaluate directly.

Factorials in Sequences

Factorials are products of all positive integers up to a certain number and appear commonly in sequences and series. In our sequence \( a_{n} = \frac{36^n}{n!} \), the factorial appears in the denominator. This has specific implications for the behavior of the sequence:

• **Growth of Factorials:** Factorials grow extremely rapidly compared to exponential functions. For instance, 3 factorial or \(3!\) equals 6, while 5 factorial or \(5!\) equals 120, and the numbers keep getting larger quickly.
• **Impact on Convergence:** Because of their rapid growth, factorials in the denominator can significantly dampen the growth of sequences. That's why sequences or series with factorial terms tend to converge, especially when the numerator does not grow as fast.

Having factorials in sequences helps manage growth, and when combined with the Ratio Test, we can predict and confirm convergence effectively. In our exercise, even though \(36^n\) grows exponentially, \(n!\) in the denominator takes over, pushing the sequence to converge towards 0 as \(n\) increases.