Solutions to Homework 6

Problem 2.

Since $\sqrt{n^2+n}-n=\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}$ we can compute the limit as

$$\lim_{n\to\infty}\frac1{\sqrt{1+1/n}+1}=1.$$ (25)

Problem 7.

By the Cauchy-Schwarz inequality,

$$\left(\sum\limits_{n=1}^N\frac{\sqrt{a_n}}n\right)^2\le \sum\limits_{n=1}^N\frac1{n^2}\sum\limits_{n=1}^Na_n.$$ (26)

Both series on the right are convergent, hence the partial sums are bounded so the partial sum on the left is bounded, hence, being a series of non-negative terms, convergent.

Problem 12.

(a) Since the $a_n>0,$ $r_n$ is strictly decreasing as $n$ increases. Thus for $m<n,$

$$\frac{a_m}{r_m}+\dots+\frac{a_n}{r_n}>\frac1{r_m}\left(a_m+\dots a_n\right)=\frac{r_n-r_m}{r_m}=1-\frac{r_n}{r_m}.$$ (27)

It follows that the series $\sum\limits_{n}\frac{a_n}{r_n}$ is not Cauchy since the right side tends to $1$ as $n\to\infty$ for fixed $m.$ Thus the series does not converge.

(b) Using the identity $(\sqrt{r_n}-\sqrt{r_{n+1}})(\sqrt{r_n}-\sqrt{r_{n+1}})=r_n-r_{n+1}=a_n$ and the fact that $r_n$ is strictly decreasing, we conclude that

$$a_n<2\sqrt{r_n}(\sqrt{r_n}-\sqrt{r_{n+1}})$$ (28)

giving the desired estimate. From this inequality we find that

$$\sum\limits_{n=1}^q\frac{a_n}{\sqrt{r_n}}<\sqrt{r_1}-\sqrt{r_{p+1}}<\sqrt{r_1}$$ (29)

so this series with positive terms is bounded and hence convergent.

Problem 16.

(a) Proceeding inductively we can assume (since it is true for $n=1)$ that $x_n>\sqrt\alpha.$ Then $x^2_n-2\sqrt\alpha x_n+\alpha=(x-\sqrt\alpha )^2>0$ so $x^2_n+\alpha>2x_n\sqrt\alpha$ and hence

$$x_{n+1}=\frac12\left(x_n+\frac\alpha {x_n}\right)>\sqrt\alpha .$$

Since $\alpha /x_n<x_n$ also follows that $x_{n+1}< x_n$ so the sequence is strictly decreasing but always larger than $\sqrt\alpha.$ Thus the limit $x_n\to x\ge\sqrt\alpha$ exists. Since $2x_{n}x_{n+1}=x_n^2-\alpha$ the limit must satisfy $2x^2=x^2-\alpha,$ that is $x=\sqrt\alpha.$

(b)Defining $\epsilon _n=x_n-\sqrt\alpha$ we find that

$$\epsilon _{n+1}=\frac1{2x_n}\left(x_n^2-2x_n\sqrt\alpha+\alpha\right)= \frac{\epsilon _n^2}{2x_n}<\frac{\epsilon _n^2}{2\sqrt\alpha}.$$

Since this is true for all $n,$ if we set $\gamma _n=\epsilon _n/\beta ,$ where $\beta=2\sqrt\alpha$ then

$$\gamma _{n+1}<\gamma _n^2\Longrightarrow \gamma _{n+1}<\gamma _1^{2^n},$$

so $\epsilon _{n+1}<\beta\left(\frac{\epsilon _1}{\beta }\right)^{2^n}.$

(c) If $\alpha =3$ and $x_1=2$ then $1\frac7{10}<\sqrt3<1\frac8{10}$ so $\epsilon _1=2-\sqrt3<\frac2{10},$ $2\sqrt3>2$ and $\epsilon _1/\beta <\frac1{10}.$ Since $\beta <4,$ $\epsilon _5<4\cdot{10}^{-16}$ and $\epsilon _6<4\dot {10}^{-32}.$

Problem 20.

Suppose thet $\{p_n\}$ is a Cauchy sequence and some subsequence $\{p_{n(k)}\}$ converges to $p.$ Then, given $\epsilon >0$ there exists $N$ such that for $n,m\ge N$ $d(p_n,p_m)<\epsilon /2$ and there exists $N'$ such that $k>N'$ implies $d(p,p_{n(k)})<\epsilon /2.$ We can choose $k>N'$ so large that $n(k)>N$ and then

$$d(p,p_n)\le d(p,p_{n(k)})+d(p_n,p_{n(k)})<\epsilon /2+\epsilon /2=\epsilon$$

provided only that $n\ge N.$ Thus $p_n\to p.$

Problem 21.

If $\{E_n\}$ is a decreasing sequence of non-empty closed sets in a metric space then there is a sequence $\{p_n\}$ with $p_n\in E_n.$ The assumption that $\operatorname{diam}E_n\to0$ means that given $\epsilon >0$ there exists $N$ such that $n\ge N$ implies $d(p,q)<\epsilon$ if $p,q\in E_n.$ Now, for $n\ge m\ge N,$ $p_n\in E_n\subset E_m$ so $d(p_n,p_m)<\epsilon.$ It follows that the sequence is Cauchy and hence, by the assumed completeness of $X$ that it converges to $p.$ Since the sequence is in $E_n$ for $m\ge n,$ $p\in E_n$ for all $n$ so $p\in\bigcap_nE_n$ as desired. Conversely there is only one point in this set since $q\in \bigcap_nE_n$ implies $d(p,q)\le\operatorname{diam}(E_n)\to0$ so $p=q.$