Chapter 9: Problem 113
Determine if the sequence is monotonic and if it is bounded. $$a_{n}=\frac{2^{n} 3^{n}}{n !}$$
Short Answer
Expert verified
The sequence is monotonic (decreasing for \( n \geq 6 \)) and bounded.
Step by step solution
01
Understanding the Sequence
We are given the sequence \( a_n = \frac{2^n 3^n}{n!} \). To understand how it behaves, note that the sequence is derived by dividing the product of powers of 2 and 3 by the factorial of \( n \). This suggests that as \( n \) increases, the factorial in the denominator grows quickly compared to the product in the numerator.
02
Analyze Monotonicity
To determine if the sequence \( a_n \) is monotonic, we consider the ratio \( \frac{a_{n+1}}{a_n} \). This gives us:\[\frac{a_{n+1}}{a_n} = \frac{2^{n+1} 3^{n+1}/(n+1)!}{2^n 3^n/n!} = \frac{6}{n+1}. \]If \( \frac{6}{n+1} < 1 \) for all \( n \geq 6 \), \( a_{n+1} < a_n \), meaning that the sequence is decreasing for \( n \geq 6 \). Thus, the sequence is not increasing but decreasing beyond a certain point.
03
Analyze Boundedness
A sequence is bounded if it has an upper and lower limit. Since \( a_n = \frac{2^n 3^n}{n!} \), we recall that \( n! \) grows faster than exponentially. For sufficiently large \( n \), \( n! > 6^n \), hence \( a_n \rightarrow 0 \) as \( n \rightarrow \infty \). This implies that the sequence is bounded below by 0. As \( a_n \geq 0 \) for all \( n \), and \( a_n \rightarrow 0 \), the sequence is bounded both above and below.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Monotonic Sequence
A sequence is monotonic if it consistently increases or decreases. To check if a sequence is monotonic, we often examine consecutive terms. For the sequence \(a_n = \frac{2^n 3^n}{n!}\), we compared the ratio of consecutive terms: \[\frac{a_{n+1}}{a_n} = \frac{6}{n+1}.\] Notice that this fraction becomes less than 1 when \( n \geq 6 \). When this happens, each term becomes smaller than the previous one, meaning the sequence decreases at least from this point onwards.
- If a sequence is continuously increasing, it is called an increasing sequence.
- If a sequence is continuously decreasing, it is called a decreasing sequence.
- Since our sequence decreases after a certain point, it acts as a decreasing sequence beyond \( n = 6 \).
Bounded Sequence
A sequence is considered bounded if all its terms stay within a fixed range. For the sequence \(a_n = \frac{2^n 3^n}{n!}\), we noted this feature by observing its behavior as \( n \to \infty \). The key observation here is the role of \( n! \), the factorial in the denominator.
- The factorial grows very rapidly, outpacing exponential growth like \( 6^n \).
- As \( n \) becomes large, \( n! \) becomes significantly larger than \( 2^n 3^n \), causing \( a_n \) to shrink towards 0.
- Therefore, the sequence is bounded above, as \( a_n \leq \frac{2^n 3^n}{6^n} = 1 \) for large \( n \).
- Since \( a_n \geq 0 \) naturally, it is also bounded below.
In short, \( a_n \) is squeezed between 0 and an upper bound close to 1, keeping it nicely bounded.
- The factorial grows very rapidly, outpacing exponential growth like \( 6^n \).
- As \( n \) becomes large, \( n! \) becomes significantly larger than \( 2^n 3^n \), causing \( a_n \) to shrink towards 0.
- Therefore, the sequence is bounded above, as \( a_n \leq \frac{2^n 3^n}{6^n} = 1 \) for large \( n \).
- Since \( a_n \geq 0 \) naturally, it is also bounded below.
In short, \( a_n \) is squeezed between 0 and an upper bound close to 1, keeping it nicely bounded.
Factorials in Sequences
Factorials like \( n! \) can significantly influence the growth or decline of a sequence. A factorial is the product of all integers up to \( n \), making the number grow remarkably fast as \( n \) grows.
For the sequence \(a_n = \frac{2^n 3^n}{n!}\):
For the sequence \(a_n = \frac{2^n 3^n}{n!}\):
- The numerator \(2^n 3^n\) rises exponentially, but
- The denominator \(n!\) rises even faster.
- Factorials are useful in determining limits and convergence of sequences.
- In this example, the factorial ensures that \( a_n \) remains bounded and approaches 0.