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

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

Problem 7.
By the Cauchy-Schwarz inequality,

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

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

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

giving the desired estimate. From this inequality we find that

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

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

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

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

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

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

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

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

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.$

Richard B. Melrose 2004-05-18