Exercise 3.17

(a) First we show that $x_n>\sqrt{\alpha }$ implies that $0<x_{n+1}<\sqrt{\alpha }$. Since clearly $\sqrt{\alpha }>1$ (since $\alpha >1$) we have

Identical reasoning with the direction reversed shows that $0<x_n<\sqrt{\alpha }$ implies that $x_{n+1}>\sqrt{\alpha }$. It follows immediately from these that $x_n>\sqrt{\alpha }$ if $n$ is odd and $0<x_n<\sqrt{\alpha }$ if $n$ is even.

Now for any $n\ge 1$ we have

So supposing that $n$ is odd we have

Thus we have shown that the subsequence of odd indices is monotonically decreasing as required.

(b) An identical derivation as that above with the direction reversed shows that the subsequence of even indices is monotonically increasing.

(c) Consider the subsequence of odd indices, i.e. $\{x_{k_n}\}$ where $k_n=2n-1$ for $n\ge 1$. We have shown above that this subsequence is bounded (above by $x_1$ and below by $\sqrt{\alpha }$) and is monotonic. Therefore by Theorem 3.14 it converges. So let $x=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{x}_{k_n}$. Then also clearly $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{x}_{k_{n+1}}=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{x}_{k_n}=x$. So since $k_{n+1}=2(n+1)-1=2n+1=(2n-1)+2=k_n+2$ by (1) we have

Solving this for $x$ results in $x=\sqrt{\alpha }$. A similar argument for the subsequence of even indices, i.e. $\{x_{l_n}\}$ where $l_n=2n$ for $n\ge 1$, shows that it also converges to $\sqrt{\alpha }$.

So consider any $𝜖>0$. Then there is an $N$ where $|x_{k_n}-\sqrt{\alpha }|<𝜖$ for all $n\ge N$. Likewise there is an $M$ where $|x_{l_n}-\sqrt{\alpha }|<𝜖$ for all $n\ge M$. So let $L=\max <!--\; FUNCTION\; APPLICATION\; -->(k_N,l_M)$ and consider any $n\ge L$. If $n$ is odd then there is an $m\ge 1$ where $n=2m-1=k_m$. We then have

so that $|x_n-\sqrt{\alpha }|=|x_{k_m}-\sqrt{\alpha }|<𝜖$. A similar argument shows that $|x_n-\sqrt{\alpha }|<𝜖$ in the case when $n$ is even as well. Thus we have shown that $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{x}_n=\sqrt{\alpha }$ by definition.

(d)

Lemma 3.17.1: If $s_n\to 0$ for a real sequence $\{s_n\}$ then also

Proof: Consider any $𝜖>0$. If $𝜖\ge 1$ then is an $N\ge 1$ such that $|s_n|<1$ for any $n\ge N$. So consider any $n\ge N$. We then have

If $𝜖<1$ then ${𝜖}^n\le 𝜖$ for $n\ge 1$. Also there is an $N\ge 1$ such that $|s_n|<𝜖$ for all $n\ge N$. So consider any $n\ge N$, nothing that also $n\ge N\ge 1$ so that

So since $𝜖$ was arbitrary we have shown that ${(s_n)}^n\to 0$. □

Lemma 3.17.2: For $0<\xi <1$ and positive real $a$

Proof: First consider any $n\ge 2$. Then by the Binomial Theorem we have

Since $\xi <1$ and $2^n>n(n-1)∕2$ it follows that for $n\ge 2$

But we have

and clearly, since $|\xi |<1$, by Theorem 3.20e

so that by Lemma 3.17.1

The desired result then follows from (2), (3), and Theorem 3.19. □

To start the main argument let $\{y_n\}$ denote the sequence from Exercise 3.16 and $\{x_n\}$ denote the sequence from this exercise, setting $y_1=x_1>\sqrt{\alpha }$. We also define ${\delta }_n=y_n-\sqrt{\alpha }$ and ${𝜖}_n=|x_n-\sqrt{\alpha }|$, noting that the absolute value for ${𝜖}_n$ is necessary since it was shown in part a that $x_n<\sqrt{\alpha }$ for even $n$.

First we show that ${\delta }_n\to 0$, which is straightforward given that it was shown in Exercise 3.16a that $y_n\to \sqrt{\alpha }$:

Next we show that there is an $N\ge 1$ where ${\delta }_N∕\beta <1$, where again $\beta =2\sqrt{\alpha }$. This follows immediately from the fact that ${\delta }_n\to 0$ since this implies that there is an $N\ge 1$ such that $|{\delta }_n-0|=|{\delta }_n|={\delta }_n<\beta$ for all $n\ge N$, since $\beta >0$. In particular ${\delta }_N<\beta$, from which it immediately follows that ${\delta }_N∕\beta <1$ since $\beta >0$.

Now in Exercise 3.16b it was shown that ${\delta }_{n+1}<{\delta }_n^2∕\beta$ for all $n\ge 1$. Similar to the derivation that follows in Exercise 3.16b we claim that ${\delta }_n\le \beta {({\delta }_N∕\beta )}^{2^{n-N}}$ for all $n\ge N$, and to the surprise of no one we show this by induction. For $n=N$ we have

Now assume that ${\delta }_n\le \beta {({\delta }_N∕\beta )}^{2^{n-N}}$. We then have

Turning our attention to $\{x_n\}$, we first claim that ${𝜖}_1>{𝜖}_2$. To see this we have

We also claim that ${𝜖}_{n+1}\ge {𝜖}_n\gamma$ for any $n\ge 1$, where we have let

To see this consider any $n\ge 1$. If $n$ is odd then we have

where we have used the fact that $x_n\le x_1$ for all odd $n\ge 1$ since the odd indexed sequence decreases monotonically as shown in part a. On the other hand if $n$ is even then

where we have used the fact that $x_n<\sqrt{\alpha }<x_1$ for even $n$ as shown in part a. So from this we can easily show that ${𝜖}_n\ge {𝜖}_1{\gamma }^{n-1}$ for all $n\ge 1$ by induction. For $n=1$ we have

Now assume that ${𝜖}_n\ge {𝜖}_1{\gamma }^{n-1}$. Then we have

Since $0<{\delta }_N∕\beta <1$ and $\gamma >0$, Lemma 3.17.2 implies that there is an $M\ge 1$ where

for all $m\ge M$, noting that the right hand side is positive. So then let $L=N+M$ and consider an arbitrary $n\ge L$. Letting $m=n-N$ we clearly have $n=m+N$ so that

Therefore we have

But since $n\ge L=N+M\ge N\ge 1$, by what was shown before and transitivity we have

Thus we have shown that for all $n\ge L$

This shows that $\{y_n\}$ converges more rapidly than $\{x_n\}$.