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Chapter 3: Problem 6

Let \(a>0\) and let \(z_{1}>0\). Define \(z_{n+1}:=\sqrt{a+z_{n}}\) for \(n \in \mathbb{N}\). Show that \(\left(z_{n}\right)\) converges and find the limit.

Short Answer

Expert verified
The sequence \(z_{n}\) converges and the limit is \(L = \frac{1 + \sqrt{1 + 4a}}{2}\)

Step by step solution

01

Verify Monotonicity

The sequence \((z_{n})\) is defined recursively: each term depends on the previous one. Based on the condition given, \(z_{1}>0\), you could observe from the terms that the sequence \(z_{n + 1} = \sqrt{a+z_{n}}\) is increasing, because for each term \(a+z_{n}\) is greater than \(z_{n}\), thus the square root of the higher value \(a+z_{n}\) would be bigger too. Therefore, \(z_{n + 1} \le \sqrt{a+z_{n}} \le \sqrt{a+z_{n+1}} = z_{n + 2}\), so \(z_{n+1}\) is a monotonically increasing sequence.
02

Verify Boundedness

To show boundedness, assume there exists a limit \(L\) for the sequence \(z_{n}\). Then as \(n\) approaches infinity, \(z_{n+1}\) is approximated by \(L\), that is \(z_{n+1} \approx L\). Hence, \(L = \sqrt{a+L}\). Solving this equation for \(L\) gives \(L^{2} - L - a = 0\). By quadratic formula, \(L = \frac{1 \pm \sqrt{1 + 4a}}{2}\). Since \(L\) is positive, then \(L = \frac{1 + \sqrt{1 + 4a}}{2}\) that becomes upper bound.
03

Establish Convergence

Based on steps 1 and 2, the sequence \(z_{n}\) is monotonic (increasing) and bounded (from above), using Monotone Convergence Theorem, it can be concluded that the sequence \(z_{n}\) is convergent.
04

Derive the Limit

Using the relation that \(L = \sqrt{a+L}\) which was proved before; square both sides to get rid of the square root, rearrange the terms to one side and solve the quadratic equation for \(L\), hence the limit of \(z_{n}\) exists and is equal to \(L = \frac{1 + \sqrt{1 + 4a}}{2}\).

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Most popular questions from this chapter

Prove that if \(\lim \left(x_{n}\right)=x\) and if \(x>0\), then there exists a natural number \(M\) such that \(x_{n}>0\) for all \(n \geq M\).

Show directly from the definition that the following are not Cauchy sequences. (a) \(\left((-1)^{n}\right)\), (b) \(\left(n+\frac{(-1)^{n}}{n}\right)\), (c) \((\ln n)\).

By using partial fractions, show that (a) \(\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)}=1\), (b) \(\sum_{n=0}^{\infty} \frac{1}{(\alpha+n)(\alpha+n+1)}=\frac{1}{\alpha}>0\), if \(\alpha>0\). (c) \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}=\frac{1}{4}\).

The first few terms of a sequence \(\left(x_{n}\right)\) are given below. Assuming that the "natural pattern" indicated by these terms persists, give a formula for the \(n\) th term \(x_{n}\) (a) \(5,7,9,11, \cdots\), (b) \(1 / 2,-1 / 4,1 / 8,-1 / 16, \cdots\) (c) \(1 / 2,2 / 3,3 / 4,4 / 5, \cdots\), (d) \(1,4,9,16, \cdots\).

If \(\sum x_{n}\) and \(\sum y_{n}\) are convergent, show that \(\sum\left(x_{n}+y_{n}\right)\) is convergent.
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