/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Show that \((n !)^{1 / n} \right... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 11

Show that \((n !)^{1 / n} \rightarrow \infty\).

Short Answer

Expert verified
The problem \((n!)^{1/n} \rightarrow \infty\) can be solved by first taking the natural logarithm of \((n!)^{1/n}\) to get \(\frac{1}{n}\ln(n!)\). After expanding this to \(\frac{1}{n}\left[\ln(1) + \ln(2) + \ldots + \ln(n)\right]\), we can rewrite it as \(\frac{1}{n} a_n\) where \(a_n = \int_{1}^{n}\ln(x)\,dx\). Then, \(\lim_{n\to\infty} \left[ \frac{1}{n} a_n \right] = \infty\), using that \(\lim_{n\to\infty} (1 + a_{n})^{n} = e\). Finally, since \(\lim_{n\to\infty} (n!)^{1/n} = \lim_{n\to\infty} (e^{a_n})^{1/n}\), we conclude that \((n!)^{1/n} \rightarrow \infty\).

Step by step solution

01

Taking natural logarithm of both sides

Let's first take the natural logarithm of \((n!)^{1/n}\): \[\ln\left((n!)^{1/n}\right)\] Remember the following properties of logarithms: 1. \(\ln(a^{b}) = b\ln(a)\) 2. \(\ln(a \cdot b) = \ln(a) + \ln(b)\) Using these properties, we can rewrite the expression as: \[\frac{1}{n}\ln(n!)\] Now, let's expand the logarithm of the factorial to a sum of logarithms: \[\frac{1}{n}\left[\ln(1) + \ln(2) + \ln(3) + \ldots + \ln(n)\right] \]
02

Change the expression to the form \((1 + a_n)^n\)

We can rewrite this expression as an integral, making use of the property that the integral of a function is the area under the curve of that function. Let \[a_n = \int_{1}^{n}\ln(x)\,dx\] Then, \[\frac{1}{n} \left[\ln(1) + \ln(2) + \ln(3) + \ldots + \ln(n)\right] = \frac{1}{n} a_n\] Now, let us differentiate \(a_n\) with respect to n: \[\frac{d}{dn}a_n = \frac{d}{dn}\int_{1}^{n}\ln(x)\,dx = \ln(n)\] As \(a_n\) goes to infinity, \(n\) also goes to infinity, which implies that \(\ln(n)\) goes to infinity. Thus, \[\lim_{n\to\infty} \left[ \frac{1}{n} a_n \right] = \infty\] Now, using the fact that \(\lim_{n\to\infty} (1 + a_{n})^{n} = e\): \[\lim_{n\to\infty} (e^{a_n})^{1/n} = \lim_{n\to\infty} e^{a_n/n} = e^{\infty} = \infty\]
03

Rewrite \((n!)^{1/n}\) and conclude

Therefore, \(\lim_{n\to\infty} (e^{a_n})^{1/n} = \infty\). Since this is equivalent to \(\lim_{n\to\infty} (n!)^{1/n}\), we can conclude that: \[(n!)^{1/n} \rightarrow \infty\]

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 91影视!

Key Concepts

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

Factorials
Factorials are a fundamental concept in mathematics, often used in permutations and combinations. When we say "n factorial," denoted by \(n!\), it refers to the product of all positive integers up to \(n\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). This recursive multiplication pattern is the core characteristic of factorials.

Factorials grow very rapidly as \(n\) increases. As part of the problem, we're looking into the asymptotic behavior of \((n!)^{1/n}\), suggesting that as \(n\) becomes infinitely large, \((n!)^{1/n}\) tends towards infinity. This means that the "average value" of terms in the factorial grows indefinitely. This observation is crucial for approximation and analysis in various domains, such as Stirling's approximation and probability theory.
Limits
A limit in mathematics is a value that a function approaches as the input approaches some value. In our exercise, we are specifically interested in limits as \(n\) approaches infinity. When we write \(\lim_{n \to \infty} (n!)^{1/n}\), we are examining how this expression behaves as \(n\) increases without bound.

The concept of a limit is foundational for calculus. It's essential for understanding behaviors and trends of functions at points that are not necessarily within the function's domain. When analyzing \((n!)^{1/n}\), the limit shows that the expression indeed diverges to infinity. This indicates that there is no upper bound, and it will grow endlessly, aligning with the rapid growth characteristic of factorials.
Logarithms
Logarithms are powerful mathematical tools that allow us to transform multiplicative relationships into additive ones, making them easier to solve. A logarithm answers the question: "To what power do we need to raise a specific base to obtain a certain number?" For example, when using natural logarithms, which have the base \(e\), \(\ln(e^x) = x\).

In our step-by-step solution, logarithms play a key role in simplifying the expression \((n!)^{1/n}\). By applying the property \(\ln(a^b) = b\ln(a)\), we're able to break down and manage the factorial inside the limit. This transforms it into a sum of logarithms, which are easier to integrate and differentiate. The transformation allows us to analyze behavior at infinity, ultimately proving that \((n!)^{1/n}\) grows boundlessly. This illustrates how logarithms can simplify complex algebraic operations and aid in solving problems related to asymptotic analysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(A_{n}:=1+\left(1 / 2^{2}\right)+\cdots+\left(1 / n^{2}\right)\) for \(n \in \mathbb{N}\). Show that there is no real number \(\alpha<1\) such that \(\left|A_{n+1}-A_{n}\right| \leq \alpha\left|A_{n}-A_{n-1}\right|\) for all \(n \in \mathbb{N}\) with \(n \geq 2\), but \(\left(A_{n}\right)\) is a Cauchy sequence.

Let \(\left(a_{n}\right)\) be a sequence in \(\mathbb{R}\) and let \(\left(b_{n}\right)\) be the sequence of arithmetic means of \(\left(a_{n}\right)\). Show that if \(a_{n} \rightarrow \infty\), then \(b_{n} \rightarrow \infty\), and also that if \(a_{n} \rightarrow-\infty\), then \(b_{n} \rightarrow-\infty\). Further, show that the converse is not true.

Let \(\left(a_{n}\right)\) and \(\left(b_{n}\right)\) be sequences in \(\mathbb{R}\). Under which of the following conditions is the sequence \(\left(a_{n} b_{n}\right)\) convergent? Justify. (i) \(\left(a_{n}\right)\) is convergent. (ii) \(\left(a_{n}\right)\) is convergent and \(\left(b_{n}\right)\) is bounded. (iii) \(\left(a_{n}\right)\) converges to 0 and \(\left(b_{n}\right)\) is bounded. (iv) \(\left(a_{n}\right)\) and \(\left(b_{n}\right)\) are convergent.

Let \(a_{n}:=1 / 2+1 / 4+\cdots+1 /(2 n)\) and \(b_{n}:=1+1 / 3+\cdots+1 /(2 n-1)\) for \(n \in \mathbb{N}\). Show that \(a_{n} \rightarrow \infty\) and \(b_{n} \rightarrow \infty\). (Hint: Example \(2.10\) (iii).)

Show that the set \([0,1)\) is uncountable. Deduce that \(\mathbb{R}\) is uncountable. (Hint: If \(f: \mathbb{N} \rightarrow[0,1)\) is bijective, then consider \(y \in[0,1)\) whose \(n\) th digit is 1 or 0 according as the \(n\) th digit of \(f(n)\) is zero or nonzero.)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.