Chapter 10: Problem 83
Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\frac{3^{n} \cdot 6^{n}}{2^{-n} \cdot n !} $$
Short Answer
Expert verified
The sequence converges with a limit of 0.
Step by step solution
01
Simplify the Sequence
Firstly, analyze the given sequence: \(a_n = \frac{3^n \cdot 6^n}{2^{-n} \cdot n!}\). Simplify this expression by rewriting it:\[a_n = \frac{(3 \cdot 6)^n}{2^{-n} \cdot n!} = \frac{18^n}{\frac{1}{2^n} \cdot n!} = \frac{18^n \cdot 2^n}{n!}\].This gives us the simplified form \(a_n = \frac{(36)^n}{n!}\).
02
Analyze the Terms as n Grows
Now, consider the behavior of \(a_n = \frac{36^n}{n!}\) as \(n\) approaches infinity. The key is to compare the growth of both the numerator \(36^n\) and the denominator \(n!\).
03
Apply the Ratio Test
To determine the convergence, apply the Ratio Test. Compute the limit:\[\lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{{n \to \infty}} \left| \frac{36^{n+1}}{(n+1)!} \cdot \frac{n!}{36^n} \right| = \lim_{{n \to \infty}} \frac{36 \cdot 36^n \cdot n!}{36^n \cdot (n+1) \cdot n!}\]This simplifies to:\[\lim_{{n \to \infty}} \frac{36}{n+1} = 0\]Since the limit is 0, which is less than 1, the sequence converges.
04
Identify the limit of the sequence
As derived from Step 3, since the sequence \(a_n\) converges and the terms become negligible as \(n\) increases, the sequence's limit is 0.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ratio Test
The Ratio Test is a handy tool for determining the convergence of sequences and series. It's particularly useful when dealing with sequences involving factorials and exponential terms. Here's how it works — for a sequence defined as:
- \( a_n \)
- \( \frac{a_{n+1}}{a_n} \)
- \[ \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \]
- If the limit is less than 1, the sequence converges.
- If the limit is greater than 1, or if it diverges to infinity, the sequence diverges.
- If the limit equals 1, the test is inconclusive.
Factorial Growth
Factorials grow extremely fast compared to exponential terms. To understand this, note that a factorial \( n! \) is the product of all positive integers from 1 to \( n \). As \( n \) increases, this product increases rapidly.Consider how exponential growth, such as \( 36^n \), behaves. Each term is simply 36 times the previous term, but for factorials, each new term rapidly multiplies by an increasing integer.For example:
- For \( n = 5 \), the factorial \( 5! = 1 \times 2 \times 3 \times 4 \times 5 = 120 \).
- For \( n = 6 \), this becomes \( 6 \times 120 = 720 \), a much faster growth.
Limit of Sequence
Finding the limit of a sequence often tells us what value the terms are approaching as \( n \) becomes very large. If a sequence converges, it means that the terms will get closer and closer to a specific number.Take our sequence:
- \( a_n = \frac{36^n}{n!} \)