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Chapter 3: Problem 14

Show that if \(z_{n}:=\left(a^{n}+b^{n}\right)^{1 / n}\) where \(0

Short Answer

Expert verified
Therefore, using the concept of sequences and series, we have successfully shown that the sequence \(z_n\) with \(z_{n}\) defined as \( \left(a^{n}+b^{n}\right)^{1 / n}\) where \(0<a<b\), converges to \(b\), i.e., \(\lim \left(z_{n}\right)=b\).

Step by step solution

01

Sequence Setup

We are given a sequence \(z_{n}\) where \(z_{n}\) is defined as \( \left(a^{n}+b^{n}\right)^{1 / n}\). Here, \(a\) and \(b\) are constants such that \(0<a<b\). We need to show that this sequence converges to \(b\) as \(n\)tends to infinity.
02

Apply binomial theorem

We will show that \(a^{n} + b^{n}\) can always be written as \(b^{n}\) multiplied by a binomial series. Therefore, \(a^{n} + b^{n} = b^{n} \left(1 + \left(\frac{a}{b}\right)^n\right)\). Taking the \(n\)th root gives \( \left(a^{n}+b^{n}\right)^{1 / n} = b \left(1 + \left(\frac{a}{b}\right)^n\right)^{1 / n}\). This is equivalent to our sequence \(z_{n} = b \left(1 + \left(\frac{a}{b}\right)^n\right)^{1 / n}\).
03

Show inequality

We can see that a/b < 1. Raise it to the power \(n\) and it will converge to \(0\) as \(n\) becomes infinitely large so that becomes \(0 that < (a/b)^{n} < 1\).
04

Convergence of limit

Substituting this into step 2 gives us \(b \leq z_{n} < b \left(1 + 1\right)^{1 / n}\). Notice that as \(n\) becomes infinitely large, \(\left(1 + 1\right)^{1 / n}\) will become \(1\). Therefore, by the sandwich theorem, \(z_{n}\) will converge to \(b\) as \(n\) tends to infinity. Hence, \(\lim \left(z_{n}\right) = b\).

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Most popular questions from this chapter

Show that if \(x_{n} \geq 0\) for all \(n \in \mathbb{N}\) and \(\lim \left(x_{n}\right)=0\), then \(\lim \left(\sqrt{x_{n}}\right)=0\).

Let \(\sum_{n=1}^{\infty} a(n)\) be such that \((a(n))\) is a decreasing sequence of strictly positive numbers. If \(s(n)\) denotes the \(n\) th partial sum, show (by grouping the terms in \(s\left(2^{n}\right)\) in two different ways) that \(\frac{1}{2}\left(a(1)+2 a(2)+\cdots+2^{n} a\left(2^{n}\right)\right) \leq s\left(2^{n}\right) \leq\left(a(1)+2 a(2)+\cdots+2^{n-1} a\left(2^{n-1}\right)\right)+a\left(2^{n}\right)\) Use these inequalities to show that \(\sum_{n=1}^{\infty} a(n)\) converges if and only if \(\sum_{n=1}^{\infty} 2^{n} a\left(2^{n}\right)\) converges. This result is often called the Cauchy Condensation Test; it is very powerful.

Let \(\sum a_{n}\) be a given series and let \(\sum b_{n}\) be the series in which the terms are the same and in the same order as in \(\sum a_{n}\) except that the terms for which \(a_{n}=0\) have been omitted. Show that \(\sum a_{n}\) converges to \(A\) if and only if \(\sum b_{n}\) converges to \(A\).

Suppose that \(x_{n} \geq 0\) for all \(n \in \mathbb{N}\) and that \(\lim \left((-1)^{n} x_{n}\right)\) exists. Show that \(\left(x_{n}\right)\) converges.

Determine the limits of the following. (a) \(\left((3 n)^{1 / 2 n}\right)\), (b) \(\left((1+1 / 2 n)^{3 n}\right)\).
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