/*! 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 7 For \(n \in \mathbb{N}\), let \(... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 7

For \(n \in \mathbb{N}\), let \(a_{n}:=n^{1 / n} .\) Show that \(a_{1}a_{n+1}\) for all \(n \geq 3\). Use this to show that the sequence \(\left(a_{n}\right)\) is convergent.

Short Answer

Expert verified
We need to prove that \(a_1 < a_2 < a_3\), and \(a_n > a_{n+1}\) for all \(n \geq 3\). After differentiation of \(f(x) = x^{1/x}\), we find out that the function is increasing on the interval \((0, e]\), and decreasing on the interval \([e, \infty)\). Thus, we have \(a_1 < a_2 < a_3\), and \(a_n > a_{n+1}\) for all \(n \geq 3\). The sequence \(\{a_n\}\) is strictly decreasing for \(n \geq 3\) and bounded below by 0. By the monotonic sequence theorem, this implies the sequence \(\{a_n\}\) is convergent.

Step by step solution

01

Find the first derivative of the function \(f(x) = x^{1/x}\)

We will first find the first derivative using the fact that \(e^{\ln(x)} = x\) and then differentiating implicitly: Let \(y = x^{1/x}\) and take the natural logarithm of both sides, \(\ln(y) = \ln(x^{1/x}) = \frac{1}{x}\ln(x)\). Now, differentiate both sides w.r.t. \(x\), \(\frac{dy}{dx}\frac{1}{y} = \frac{-\ln(x)+1}{x^2}\). Finally, multiply both sides by \(y\), \(\frac{dy}{dx} = x^{1/x}\left(\frac{-\ln(x)+1}{x^2}\right)\).
02

Find the critical points of the function

To find the critical points, we need to solve \(\frac{dy}{dx} = 0\). Notice that \(x^{1/x} > 0\) for all \(x>0\), so we can ignore it. Thus, \(-\ln(x)+1 = 0 \Rightarrow x = e\).
03

Determine where the function is increasing and decreasing

The function is increasing when \(\frac{dy}{dx} > 0\) and decreasing when \(\frac{dy}{dx} < 0\). If \(x < e\), then \(-\ln(x)+1 > 0\) and the function is increasing. When \(x > e\), then \(-\ln(x)+1 < 0\) and the function is decreasing. Thus, the function is increasing on the interval \((0, e]\), and decreasing on the interval \([e, \infty)\).
04

Prove that \(a_1 < a_2 < a_3\)

From step 3, we know that the function is increasing on the interval \((0, e]\). Since \(1 < 2 < 3 < e\), we have \(\displaystyle a_1 = f(1) < f(2) = a_2 < f(3) = a_3\).
05

Prove that \(a_n > a_{n+1}\) for all \(n \geq 3\)

From step 3, we know that the function is decreasing on \([e, \infty)\). Since all \(n \geq 3\) are greater than \(e\), we have \(a_n > a_{n+1}\) for all \(n \geq 3\).
06

Prove the sequence is convergent

Since \(a_1 < a_2 < a_3\) and \(a_n > a_{n+1}\) for all \(n \geq 3\), the sequence \(\{a_n\}\) is strictly decreasing for \(n \geq 3\) and bounded below by 0, as \(n^{1/n} > 0\) for all \(n\). By the monotonic sequence theorem, this implies the sequence \(\{a_n\}\) is convergent.

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Most popular questions from this chapter

Let \(x \in \mathbb{R}\) and \(x>0 .\) Define $$ A_{n}:=1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\cdots+\frac{x^{n}}{n !} \quad \text { and } \quad B_{n}:=\left(1+\frac{x}{n}\right)^{n} \quad \text { for } n \in \mathbb{N} . $$ Show that \(\left(A_{n}\right)\) and \(\left(B_{n}\right)\) are convergent and have the same limit.

Prove that a sequence \(\left(a_{n}\right)\) in \(\mathbb{R}\) has no convergent subsequence if and only if \(\left|a_{n}\right| \rightarrow \infty\).

Assuming only the algebraic and the order properties of \(\mathbb{R}\), and assuming that every Cauchy sequence in \(\mathbb{R}\) is convergent, establish the Completeness Property of \(\mathbb{R}\). (Hint: Consider \(S \subseteq \mathbb{R}\) and \(a_{n}, \alpha_{n}\) as in the hint for Exercise 2.36. Then \(\alpha_{n}-a_{n} \leq\left(\alpha_{0}-a_{0}\right) / 2^{n}\) for all \(n \in \mathbb{N} .\) )

Given any \(m \in \mathbb{N}\), show that there is a unique nonnegative integer \(k\) such that \(10^{k} \leq m<10^{k+1}\). Use Exercise \(1.35\) repeatedly to show that there are unique integers \(a_{0}, a_{1}, \ldots, a_{k}\) between 0 and 9 such that $$ m=a_{0}+a_{1}(10)+a_{2}\left(10^{2}\right)+\cdots+a_{k}\left(10^{k}\right) $$

Let \(y\) be any real number with \(0 \leq y<1\). Define sequences \(\left(b_{n}\right)\) and \(\left(y_{n}\right)\) iteratively as follows. Let \(y_{1}:=10 y\) and \(b_{1}:=\left[y_{1}\right]\), and for each \(n \in \mathbb{N}\), let \(y_{n+1}:=10\left(y_{n}-b_{n}\right)\) and \(b_{n+1}:=\left[y_{n+1}\right] .\) Show that \(b_{n}\) are integers with \(0 \leq b_{n} \leq 9\) and \(y_{n}\) are real numbers with \(0 \leq y_{n}<10\) for each \(n \in \mathbb{N}\) and moreover, $$ y=\frac{b_{1}}{10}+\frac{b_{2}}{10^{2}}+\cdots+\frac{b_{n}}{10^{n}}+\frac{y_{n+1}}{10^{n+1}} $$ Deduce that \(0 \leq y_{n+1} / 10^{n+1}<1 / 10^{n}\) for each \(n \in \mathbb{N}\), and consequently, $$ y=\lim _{n \rightarrow \infty}\left(\frac{b_{1}}{10}+\frac{b_{2}}{10^{2}}+\cdots+\frac{b_{n}}{10^{n}}\right) $$ [Note: It is customary to call \(b_{1}, b_{2}, \ldots\), the digits of \(y\) and write the above expression as \(y=0 . b_{1} b_{2} \ldots\) and call it the decimal expansion of \(y\).]
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