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Chapter 9: Problem 82

Evaluate the limit of the following sequences. $$a_{n}=\frac{4^{n}+5 n !}{n !+2^{n}}$$

Short Answer

Expert verified
Answer: As \(n \to \infty\), the limit of the sequence is 5.

Step by step solution

01

Observe the sequence expression

Let's take a closer look at the given sequence: $$a_n = \frac{4^n + 5n!}{n! + 2^n}$$ As n approaches infinity, we will compare the growth rates of the terms in the numerator and denominator.
02

Identify dominant terms in numerator and denominator

As n approaches infinity, the dominant term in the numerator is \(5n!\), as this term will grow much faster than \(4^n\). Similarly, the dominant term in the denominator is \(n!\), as this term will grow much faster than \(2^n\). We can rewrite our sequence to focus on the dominant terms: $$a_n \approx \frac{5n!}{n!}$$ To simplify, we can cancel \(n!\) in both the numerator and the denominator: $$a_n \approx 5$$ Since the other terms do not affect the limit significantly, we can now conclude:
03

Find the limit

As \(n \to \infty\): $$\lim_{n\to\infty} a_n = 5$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit of a Sequence
A sequence is a list of numbers in a specific order. Understanding the limit of a sequence is crucial for analyzing its long-term behavior. The concept of a limit in sequences refers to what value the terms of the sequence get closer to as the index grows indefinitely. Consider the given sequence:\[a_n = \frac{4^n + 5n!}{n! + 2^n}\]To find the limit, we examine how the terms behave as \(n\) approaches infinity. If the terms stabilize to converge to a single value, that value is the limit. In this case, as \(n\) becomes very large, the sequence simplifies to a form where the factorial terms significantly overpower the exponential terms, simplifying the limit to \(5\). By simplifying the sequence appropriately, we find:\[\lim_{n\to\infty} a_n = 5\]Finding the limit of a sequence often involves identifying and focusing on dominant terms, which dictates the sequence's behavior at infinity.
Factorial Growth
Factorial growth is an essential concept when analyzing sequences. The factorial notation \(n!\) indicates the product of all positive integers up to \(n\). As \(n\) grows, \(n!\) grows exceptionally fast, even more so than exponential terms like \(2^n\) or \(4^n\).In our sequence, \(5n!\) in the numerator and \(n!\) in the denominator show how factorial growth dominates. Factorials grow super-exponentially, meaning they increase rapidly compared to simple powers. Understanding this growth helps identify the sequence's prominent terms, as was done in this case by comparing \(n!\) to other components. Recognizing \(n!\) as the dominant growth factor allows for simplifying the sequence effectively, leading to a clear evaluation of the sequence's limit.
Dominant Term Analysis
Dominant term analysis is a strategic approach to simplifying sequences. By focusing on terms with the highest growth rates, you can efficiently evaluate the overall behavior of complex sequences.For the given sequence:\[a_n = \frac{4^n + 5n!}{n! + 2^n}\]Identifying \(5n!\) as the dominant term in the numerator and \(n!\) as the dominant term in the denominator simplifies our task. This analysis highlights that these terms will eventually "overpower" other terms as \(n\) increases.With dominant terms:- Numerator: \(5n!\) surpasses \(4^n\)- Denominator: \(n!\) surpasses \(2^n\)Canceling out the \(n!\) terms results in a much simpler expression. Thus, dominant term analysis is invaluable for calculating sequence limits, as it reduces complexities and allows for straightforward limit evaluation.

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Most popular questions from this chapter

A glimpse ahead to power series Use the Ratio Test to determine the values of \(x \geq 0\) for which each series converges. $$\sum_{k=1}^{\infty} \frac{x^{k}}{k^{2}}$$

Series of squares Prove that if \(\sum a_{k}\) is a convergent series of positive terms, then the series \(\Sigma a_{k}^{2}\) also converges.

Consider the series \(\sum_{k=3}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{p}},\) where \(p\) is a real number. a. For what values of \(p\) does this series converge? b. Which of the following series converges faster? Explain. $$ \sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{2}} \text { or } \sum_{k=3}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{2}} ? $$

Given that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6},\) show that \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{2}}=\frac{\pi^{2}}{12}.\) (Assume the result of Exercise 63.)

Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer \(N\) and call it \(a_{0} .\) This is the seed of a sequence. The rest of the sequence is generated as follows: For \(n=0,1,2, \ldots\) $$a_{n+1}=\left\\{\begin{array}{ll} a_{n} / 2 & \text { if } a_{n} \text { is even } \\ 3 a_{n}+1 & \text { if } a_{n} \text { is odd. } \end{array}\right.$$ However, if \(a_{n}=1\) for any \(n,\) then the sequence terminates. a. Compute the sequence that results from the seeds \(N=2,3\), \(4, \ldots, 10 .\) You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers \(N,\) the sequence terminates after a finite number of terms. b. Now define the hailstone sequence \(\left\\{H_{k}\right\\},\) which is the number of terms needed for the sequence \(\left\\{a_{n}\right\\}\) to terminate starting with a seed of \(k\). Verify that \(H_{2}=1, H_{3}=7,\) and \(H_{4}=2\). c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?
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