Problem 64
Question
Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(27-90\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\frac{(-4)^{n}}{n !} $$
Step-by-Step Solution
Verified Answer
The sequence converges to 0.
1Step 1: Analyze Behavior of Numerator and Denominator
Consider the sequence \( a_n = \frac{(-4)^n}{n!} \). The numerator \((-4)^n\) is an exponential function that grows rapidly in magnitude as \(n\) increases due to repeated multiplication of 4. Meanwhile, the denominator \(n!\) is a factorial term that grows even more rapidly because each new term is the product of all positive integers up to \( n \).
2Step 2: Compare Growth Rates Using Limits
Analyze the limit \( \lim_{{n \to \infty}} \frac{(-4)^n}{n!} \). As \(n\) approaches infinity, factorials tend to outgrow exponential terms. Using the ratio test or intuitively comparing growth rates, note that \(n!\) grows much faster than \((-4)^n\), pushing the terms \(a_n\) towards 0.
3Step 3: Apply Limit Laws
By the definition of a limit and properties of factorial growth: \[ \lim_{{n \to \infty}} \frac{(-4)^n}{n!} = 0. \] This is because the factorial in the denominator increases slightly faster than exponential growth in the numerator.
4Step 4: Conclude Sequence Behavior
Since the limit of \( \frac{(-4)^n}{n!} \) as \(n\) approaches infinity is 0, the sequence \( a_n \) converges to 0. The larger factorial in the denominator pulls the sequence towards 0 as \(n\) becomes very large.
Key Concepts
Factorial GrowthExponential GrowthLimit of a Sequence
Factorial Growth
Factorial growth refers to a type of growth where each number in a sequence is the product of all positive integers up to a certain number, denoted as \( n! \). Factorial growth is incredibly rapid because the multiplication of integers adds up quickly.
For example, the factorial of 4, written as \( 4! \), is calculated as \( 4 \times 3 \times 2 \times 1 = 24 \). As the value of \( n \) increases, so do the number of multiplications involved, leading to extremely rapid growth.
For example, the factorial of 4, written as \( 4! \), is calculated as \( 4 \times 3 \times 2 \times 1 = 24 \). As the value of \( n \) increases, so do the number of multiplications involved, leading to extremely rapid growth.
- Factorial of 5 is \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
- Factorial of 6 is \( 6! = 720 \).
Exponential Growth
Exponential growth occurs when a quantity increases rapidly by a constant multiplier. In mathematical terms, the function \((-4)^n\) grows exponentially since each subsequent term is the product of the previous term and \(-4\).
Consider:
Consider:
- \((-4)^1 = -4\)
- \((-4)^2 = 16\)
- \((-4)^3 = -64\)
Limit of a Sequence
The limit of a sequence refers to the behavior of the terms within that sequence as the index \( n \) approaches infinity. If a sequence has a limit, it converges to that limit; otherwise, it diverges.
For a sequence like \( a_n = \frac{(-4)^n}{n!} \), it is crucial to determine whether its terms approach a specific number as \( n \) gets larger. When analyzing its limit, you consider each component individually:
Applying limit laws confirms that
\[ \lim_{{n \to \infty}} \frac{(-4)^n}{n!} = 0. \]
Thus, the sequence is said to converge to 0. Knowing how to find the limit tells us a lot about how a sequence will behave in the long run.
For a sequence like \( a_n = \frac{(-4)^n}{n!} \), it is crucial to determine whether its terms approach a specific number as \( n \) gets larger. When analyzing its limit, you consider each component individually:
- The numerator \((-4)^n\) shows exponential growth.
- The denominator \(n!\) exhibits factorial growth, which is even faster.
Applying limit laws confirms that
\[ \lim_{{n \to \infty}} \frac{(-4)^n}{n!} = 0. \]
Thus, the sequence is said to converge to 0. Knowing how to find the limit tells us a lot about how a sequence will behave in the long run.
Other exercises in this chapter
Problem 64
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