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The sequence \(\\{n !\\}\) ultimately grows faster than the sequence \(\left\\{b^{n}\right\\},\) for any \(b>1,\) as \(n \rightarrow \infty .\) However, \(b^{n}\) is generally greater than \(n !\) for small values of \(n\). Use a calculator to determine the smallest value of \(n\) such that \(n !>b^{n}\) for each of the cases \(b=2, b=e,\) and \(b=10\).

Short Answer

Expert verified
Answer: For \(b=2\), the smallest value of n is 4; for \(b=e\), the smallest value of n is 4; and for \(b=10\), the smallest value of n is 8.

Step by step solution

01

Calculate Factorial and Exponential Terms

Calculate the terms for both sequences for increasing values of n, until \(n! > 2^n\). Start with n = 1 and then continue with n = 2, 3, 4, ... - When n = 1: \(1! = 1\) and \(2^1 = 2\) - When n = 2: \(2! = 2\) and \(2^2 = 4\) - When n = 3: \(3! = 6\) and \(2^3 = 8\) - When n = 4: \(4! = 24\) and \(2^4 = 16\) In this case, we find the smallest value of n = 4 where \(n! > 2^n\). #Case 2: b=e#
02

Calculate Factorial and Exponential Terms

Calculate the terms for both sequences for increasing values of n, until \(n! > e^n\). Use the value of e ≈ 2.71828. Start with n = 1 and then continue with n = 2, 3, 4, ... - When n = 1: \(1! = 1\) and \(e^1 = 2.71828\) - When n = 2: \(2! = 2\) and \(e^2 = 7.38906\) - When n = 3: \(3! = 6\) and \(e^3 = 20.08554\) - When n = 4: \(4! = 24\) and \(e^4 = 54.59815\) In this case, we find the smallest value of n = 4 where \(n! > e^n\). #Case 3: b=10#
03

Calculate Factorial and Exponential Terms

Calculate the terms for both sequences for increasing values of n, until \(n! > 10^n\). Start with n = 1 and then continue with n = 2, 3, 4, ... - When n = 1: \(1! = 1\) and \(10^1 = 10\) - When n = 2: \(2! = 2\) and \(10^2 = 100\) - When n = 3: \(3! = 6\) and \(10^3 = 1000\) - When n = 4: \(4! = 24\) and \(10^4 = 10000\) - When n = 5: \(5! = 120\) and \(10^5 = 100000\) - When n = 6: \(6! = 720\) and \(10^6 = 1000000\) - When n = 7: \(7! = 5040\) and \(10^7 = 10000000\) - When n = 8: \(8! = 40320\) and \(10^8 = 100000000\) In this case, we find the smallest value of n = 8 where \(n! > 10^n\). In conclusion, the smallest values of n for which the factorial sequence becomes larger than the exponential sequence in the given cases are: - For \(b=2\), n = 4 - For \(b=e\), n = 4 - For \(b=10\), n = 8

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Key Concepts

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

Exponential Growth
Exponential growth is a powerful mathematical concept where a quantity increases at a rate proportional to its current value. This type of growth is common in various real-world scenarios, like population growth or interest in banks. Essentially, exponential growth can seem slow initially, but it quickly becomes rapid and substantial as time goes on.
In mathematics, exponential growth is often represented by the expression \( b^n \), where \( b \) is the base, and \( n \) is the exponent. Here, \( b \) is greater than 1, which is crucial for the growth to be genuinely exponential. For instance, values like \( b=2 \), or \( b=e \) (Euler’s number approximately 2.718) are often used in problems dealing with exponential functions.
Exponential growth is integral to understanding sequence behavior, especially in comparison to other types of sequences like factorials. While factorials grow faster in the long run, exponential functions will generally exceed factorials at smaller values of \( n \). This aspect is important when tackling problems where you need to compare when a factorial exceeds an exponential sequence.
Sequence Comparison
In mathematics, comparing sequences involves examining how different sequences behave or grow as their variable, usually represented by \( n \), increases. For example, the factorial sequence \( n! \) and the exponential sequence \( b^n \) allow us to explore different growth behaviors.

Initially, for small \( n \), the exponential sequence \( b^n \) is typically larger because it accumulates rapidly due to its multiplicative nature. For example, with small numbers, \( 2^n \), \( e^n \), or \( 10^n \) can easily surpass \( n! \). However, as n increases, factorials, stacked products of numbers, grow significantly faster than exponentials, demonstrating a critical point about how sequences diverge.

In exercises comparing sequences, one might calculate individual terms, like \( n! \) and \( b^n \), to find the smallest \( n \) at which the factorial surpasses the exponential function. Despite initial appearances, the factorial sequence ultimately grows faster, illustrating an essential shift. Learning to compare these sequences can help with understanding various phenomena, including predicting when one process outpaces another.
Growth Rates
Growth rates describe how quickly a sequence or function increases as its input increases. Different mathematical expressions have characteristically different growth rates, impacting how they behave over time.
Factorials, denoted by \( n! \), showcase a particular kind of explosive growth because each term in the sequence multiplies by a new whole number. The result is that the growth rate of factorials is much faster than linear or even polynomial growth.
Comparing factorial growth with exponential growth is an interesting study. Though an exponential sequence like \( b^n \) is powerful, factorials eventually outpace it. This situation arises because a factorial grows by multiplying by each positive integer less than or equal to \( n \), rapidly becoming a larger sequence.
  • For a given base \( b>1 \), exponential sequences have consistent multiplicative growth rates dictated by the constant base, contributing to their initial dominance.
  • Meanwhile, factorials grow by adding more factors as \( n \) increases, allowing them to surpass even high exponential powers at larger \( n \). This shows the unique and potent growth rate of factorial sequences.
Understanding growth rates helps in various fields like economics, biology, and computer science, particularly in algorithms where determining efficiency is key. By mastering these concepts, students can better predict and explain sequence behaviors.

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Most popular questions from this chapter

Consider the alternating series $$ \sum_{k=1}^{\infty}(-1)^{k+1} a_{k}, \text { where } a_{k}=\left\\{\begin{array}{cl} \frac{4}{k+1}, & \text { if } k \text { is odd } \\ \frac{2}{k}, & \text { if } k \text { is even } \end{array}\right. $$ a. Write out the first ten terms of the series, group them in pairs, and show that the even partial sums of the series form the (divergent) harmonic series. b. Show that \(\lim _{k \rightarrow \infty} a_{k}=0\) c. Explain why the series diverges even though the terms of the series approach zero.

Evaluate the series \(\sum_{k=1}^{\infty}\left(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right)\) two ways. a. Use a telescoping series argument. b. Use a geometric series argument after first simplifying \(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\)

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?

The expression $$1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+}}}}.$$ where the process continues indefinitely, is called a continued fraction. a. Show that this expression can be built in steps using the recurrence relation \(a_{0}=1, a_{n+1}=1+1 / a_{n},\) for \(n=0,1,2,3, \ldots . .\) Explain why the value of the expression can be interpreted as \(\lim _{n \rightarrow \infty} a_{n},\) provided the limit exists. b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\). c. Using computation and/or graphing, estimate the limit of the sequence. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. e. Assuming the limit exists, use the same ideas to determine the value of $$a+\frac{b}{a+\frac{b}{a+\frac{b}{a+\frac{b}{a+}}}}$$ where \(a\) and \(b\) are positive real numbers.

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty} 3^{-k}$$

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