Real Analysis – Exercises II « Climbing the Mountain

Real Analysis – Exercises II

a) Given the sequence ${ \dfrac{n^2+1}{2n^2-1}}$ prove that there exists an order ${ k}$ where ${ \left | u_n - \dfrac{1}{2} \right |<10^{-3}}$ is valid.

${ \begin{aligned} \left | \dfrac{n^2+1}{2n^2-1} - \dfrac{1}{2} \right | &< 10^{-3} \\ \left | \dfrac{2n^2+2-2n^2+1}{2(2n^2-1)} \right | &< 10^{-3} \\ \left | \dfrac{3}{2(2n^2-1)} \right | &< 10^{-3} \\ \dfrac{|3|}{|2(2n^2-1)|} &< 10^{-3} \end{aligned}}$

Since ${ 2(2n^2-1)>0}$ what follows is

${ \begin{aligned} \dfrac{3}{2(2n^2-1)} &< 10^{-3} \\ 3/2 \times 10^3 &< 2n^2-1 \\ 3/4\times 10^3+1/2 &< n^2 \\ \sqrt{3/4\times 10^3 + 1/2} &< n \end{aligned}}$

So by taking ${ k > \left \lfloor \sqrt{3/4\times 10^3 + 1/2}\right \rfloor +1}$ we have the intended result.

b) Prove by definition that ${ u_n \rightarrow 1/2}$

By the definition of limit, and using a), we have ${ n > \sqrt{\dfrac{3}{4 \delta}+1/2}}$ . If we take ${ k= \left \lfloor \sqrt{\dfrac{3}{4 \delta}+1/2} \right\rfloor+1}$ the difference between ${ u_n}$ and ${ 1/2}$ is always less than ${ \delta}$ .

2. Prove that ${ \lim u_n = 0 \Leftrightarrow \lim |u_n| = 0}$

The gist of this result is that sometimes is easier to prove that the modulus of a sequence tends to zero than to prove the sequence tends to zero. Since one result implies the other we may avoid doing some boring calculations.

We say that ${ u_n \rightarrow a}$ iff ${ \forall \delta > 0 \, \exists k \in \mathbb{N}: \quad n>k \Rightarrow |u_n - a| < \delta}$

Thus ${ \lim |u_n - a| = 0}$ iff

${\forall \delta > 0\,\exists k\in\mathbb{N}:\; n > k\Rightarrow||u_n-a|-0| < \delta}$

${\Leftrightarrow \forall \delta > 0 \, \exists k \in \mathbb{N}:\; n > k \Rightarrow |u_n - a| < \delta}$

Hence with ${ a=0}$ the conditions ${ \lim u_n = 0}$ and ${ \lim |u_n| = 0}$ are indeed equivalent.

3. Calculate ${ \lim \sqrt{n+1}-\sqrt{n}}$

This limit we are interested in calculating can be seen as ${ \lim u_n - v_n}$ where ${ u_n = \sqrt{n+1}}$ and ${ v_n = \sqrt{n}}$ . We know that ${ \lim u_n = \lim \sqrt{n+1} = +\infty}$ and ${ \lim v_n = \lim \sqrt{n} = +\infty}$ .

So what we are trying to measure is how fast these sequences diverge. If this limit is ${ a \in \mathbb{R}^+}$ then ${ u_n}$ grows slightly faster, if it is ${ a \in \mathbb{R}^-}$ then it is ${ v_n}$ that grows slightly faster.

In the case of ${ \pm \infty}$ we have that ${ u_n}$ , for the ${ +}$ sign ( ${ v_n}$ for the ${ -}$ sign) tends infinitely faster to the given limit.

On with the calculations now:

${\begin{aligned} \lim \sqrt{n+1}-\sqrt{n} &= \lim \dfrac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}} \\ &= \lim \dfrac{n+1-n}{\sqrt{n+1} + \sqrt{n}} \\ &= \lim \dfrac{1}{\sqrt{n+1}+\sqrt{n}} \\ &= \lim \dfrac{1}{\sqrt{n(1+1/n)}+\sqrt{n}} \\ &= \lim \dfrac{1}{\sqrt{n}\sqrt{1+1/n}\sqrt{n}} \\ &= \lim \dfrac{1}{\sqrt{n}\left( \sqrt{1+1/n}+1 \right)} \\ &= \lim\dfrac{1}{\left( \sqrt{1+1/n}+1 \right) } \lim \dfrac{1}{\sqrt{n}} \\ &= \lim\dfrac{1}{2 \sqrt{n}} \\ &= 0 \end{aligned}}$

4. Calculate ${ \lim \left( \sqrt{n^2+n} - \sqrt{n^2+1} \right)}$

${\begin{aligned} \lim \left( \sqrt{n^2+n} - \sqrt{n^2+1} \right)&=\lim \dfrac{n^2+n-n^2-1}{\sqrt{n^2+n} + \sqrt{n^2+1}} \\ &=\lim \dfrac{n-1}{\sqrt{n^2\left(1+\frac{1}{n}\right)} + \sqrt{n^2\left(1+\frac{1}{n^2}\right)}} \\ &=\lim \dfrac{n-1}{n \sqrt{1+\frac{1}{n}} + n\sqrt{1+\frac{1}{n^2}}} \\ &=\lim \dfrac{n-1}{n\left( \sqrt{1+\frac{1}{n}} + \sqrt{1+\frac{1}{n^2}} \right)} \\ &=\lim \dfrac{n-1}{2n} \\ &=\dfrac{1}{2} \end{aligned}}$

5. Calculate ${ \lim \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2}}$ .

Let us write out a few terms of this sum so that we can gain some intuition of what’s going on

$\sum _{k=1}^n \dfrac{1}{(n+k)^2} = \dfrac{1}{(n+1)^2}+\dfrac{1}{(n+2)^2}+\cdots + \dfrac{1}{(2n)^2}$

So by making ${ n \rightarrow \infty}$ what we get is a bigger number of increasingly small numbers to add.

The result of this limit will tell us what of these two contradictory effects wins.

Since we are adding up ${ n}$ terms that get increasingly small we have

$\displaystyle \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2} \leq \dfrac{n}{(n+1)^2}$

And also

$\displaystyle \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2} \geq \dfrac{n}{4n^2}$

Hence ${ \dfrac{n}{4n^2} \leq \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2} \leq \dfrac{n}{(n+1)^2}}$ with ${ \lim \dfrac{n}{4n^2} = \lim \dfrac{n}{(n+1)^2} = 0}$

Thus ${ \lim \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2} = 0}$ also.

Hence the fact that the fractions tend to ${ \,0}$ is more relevant to the value of the limit than the the fact that we get an infinite number of fractions to add.

6. Calculate $\sum _{k=1}^n \dfrac{1}{\sqrt{n+k}}$

Now we have a similar situation to the previous exercise but this time the numbers that are being added are bigger than the previous ones. So it begs the question: What will be the limit this time?

Like previously we are summing ${ n}$ increasingly small terms and so

$\displaystyle \displaystyle \sum _{k=1}^n \dfrac{1}{\sqrt{n+k}} \geq \dfrac{n}{\sqrt{2n}}$

And

$\displaystyle \displaystyle \sum _{k=1}^n \dfrac{1}{\sqrt{n+k}} \leq \dfrac{n}{\sqrt{n+1}}$

Thus ${ \dfrac{n}{\sqrt{2n}} \leq \displaystyle \sum _{k=1}^n \dfrac{1}{\sqrt{n+k}} \leq \dfrac{n}{\sqrt{n+1}}}$ .

Since ${ \lim \dfrac{n}{\sqrt{2n}} = \lim \dfrac{1}{\sqrt{2}}\dfrac{n}{\sqrt{n}}= \lim \dfrac{1}{\sqrt{2}}\sqrt{n} = + \infty}$ and ${ \lim \dfrac{n}{\sqrt{n+1}} = \lim \dfrac{n}{\sqrt{n}}\dfrac{1}{\sqrt{1+1/n}} = \lim \sqrt{n}\dfrac{1}{\sqrt{1+1/n}} = +\infty}$ it follows that $\sum _{k=1}^n \dfrac{1}{\sqrt{n+k}} = + \infty$

This time the fact that the number of fractions to add grows without a bound is more relevant than the fact that those numbers tend to ${ \,0}$ .

7. Calculate ${ \lim \dfrac{n^n}{n!}}$

Let us do this visually:

${ n^{n-1} = n \times n \times n \ldots \times n}$ with ${ n-1}$ terms.

${ n! = 1 \times 2 \times 3 \times \ldots \times n = 2 \times 3 \times \ldots \times n}$ with ${ n-1}$ terms.

So ${ \lim \dfrac{n^n}{n!} \geq \lim \dfrac{n^n}{n^{n-1}} = + \infty}$ . Then also ${ \lim \dfrac{n^n}{n!} = +\infty}$

From this we can see that ${ n^n}$ grows to infinity faster than ${ n!}$

8. Give examples of sequences that

a) ${ u_n \rightarrow +\infty}$ and ${ v_n \rightarrow - \infty}$ : ${ u_n+v_n=0}$

${ u_n = n}$ and ${ v_n = -n}$

b) ${ u_n \rightarrow +\infty}$ and ${ v_n \rightarrow - \infty}$ : ${ u_n+v_n=10}$

${ u_n = n+10}$ and ${ v_n = -n}$

c) ${ u_n \rightarrow +\infty}$ and ${ v_n \rightarrow - \infty}$ : ${ u_n+v_n=+\infty}$

${ u_n = 2n}$ and ${ v_n = -n}$

d) ${ u_n \rightarrow +\infty}$ and ${ v_n \rightarrow - \infty}$ : ${ u_n+v_n}$ doesn’t exist.

${ u_n = n+(-1)^n}$ and ${ v_n = -n}$

e) ${ u_n \rightarrow 0}$ and ${ v_n \rightarrow \infty}$ : ${ u_n v_n = a \in \mathbb{R}}$

${ u_n = \dfrac{a}{n}}$ and ${ v_n = n}$

f) ${ u_n \rightarrow 0}$ and ${ v_n \rightarrow \infty}$ : ${ u_n v_n = 0}$

${ u_n = \dfrac{1}{n^2}}$ and ${ v_n = n}$

g) ${ u_n \rightarrow 0}$ and ${ v_n \rightarrow \infty}$ : ${ u_n v_n = +\infty}$

${ u_n = \dfrac{1}{n}}$ and ${ v_n = n^2}$

h) ${ u_n \rightarrow 0}$ and ${ v_n \rightarrow \infty}$ : ${ u_n v_n}$ doesn’t exist

${ u_n = \dfrac{\sin n}{n}}$ and ${ v_n = n}$

This entry was posted on February 20, 2009 at 5:46 pm and is filed under Basic Mathematics, Real Analysis . You can follow any responses to this entry through the RSS 2.0 feed You can leave a response, or trackback from your own site.

18 Responses to “Real Analysis – Exercises II”

Américo Tavares Says:
February 21, 2009 at 8:29 pm

It is a very good idea of yours posting detailed solutions as you do … and a huge amount of work.

In general I like the themes with black background, but in this particular case, I think that your readers could see the formulae better with black text and white (or similar) background.

My opinion is based on what I see in the blog surfer (black in white).
dweebydoofus Says:
February 21, 2009 at 10:16 pm

Hi there. First of all thanks for taking the time to comment.

And I also want to thank you for your suggestion. For a time I considered changing my wordpress theme because I feared that the formulas weren´t too reader firendly. But what I found out is that it really is a problem in the screen definitions. in the past I also tried the white background/ black symbols option but it felt like doing an even worse job.

Doing detailed solutions really is a pain but as the blog goes things that I feel that should be routine calculations won’t be detailed so much but in the start I’ll do it like this so that people can follow all steps in the reasoning. And I’ve started writing things in LateX and then what I do is copying and pasting it to wordpress so that the equations writting part (by far the slowest part) is considerably sped up.

E já agora eu também sou de Portugal! Sempre que queira contribuir com algum comentário por favor faça-o que eu sou todo ouvidos.
Américo Tavares Says:
February 22, 2009 at 1:12 am

Olá!

A minha opinião quanto ao grande trabalho que tem baseia-se na minha própria experiência desde o início de Outubro/2007.

Realmente gosto do tema Tarski do meu blog(ue).
Em alternativa, no meu blog auxiliar (Cálculos Auxiliares), cujo link se encontra no principal, poderá encontrar outro tema que também usa letras pretas em fundo branco.

Quando há texto corrido sem fórmulas (ou muito poucas) e muitas fotos, normalmente o tema que está a usar resulta muito bem.

Vou incluir o seu blog na minha lista, embora já seja grande demais.

Um bom Carnaval!
dweebydoofus Says:
February 26, 2009 at 9:38 pm

Sim sim é um bocado chato mas agora há uma nove ferramenta feita por um user do wordpress que facilita e simplifica muito a nossa tarefa: aqui

Vou experimentá-lo mas desde já digo que o resultado final fica muito bem como se poder no blog do terrytao
Américo Tavares Says:
February 26, 2009 at 11:05 pm

Obrigado pela informação.

No meu caso particular, que costumo usar o Scientific Word, que gera um ficheiro .tex, e não num editor de TeX ou LaTeX, não sei se conseguirei arranjar maneira de tirar partido deste conversor, pois penso ser necessário instalar um ambiente adequado, diferente do meu.
ateixeira Says:
February 27, 2009 at 9:53 pm

O conversor gera um ficheiro html a partir de um ficheiro .tex. Por isso acho que desde que use apenas as funcionalidades suportadas pelo LateX do wordpress (por exemplo não pode utilizar eqnarray) tudo correrá bem.
Américo Tavares Says:
February 28, 2009 at 10:38 am

Obrigado pelo seu esclarecimento. Agora vou experimentar.
Mike Says:
March 2, 2009 at 7:28 pm

Just passing by.Btw, you website have great content!
Palynka Says:
March 17, 2009 at 6:42 pm

*taps fingers impatiently*
ateixeira Says:
March 25, 2009 at 10:24 am

My good man sorry for the delay but this project is till alive. If anything I learned that my ability to make predictions on how long things will last is very bad.

Still I already have the next post all written up in paper (it’s about four pages) and half of it it’s already written in LateX. So just give me a few more days that I’ll take care of it.

As for our discussion on the reality of triangles I’ll have to take a little bit ore time. Once again sorry
Palynka Says:
April 1, 2009 at 9:20 am

No worries, I’m also slacking on my decision theory thread.
Américo Tavares Says:
December 29, 2009 at 9:09 pm

Determine os limites das sucessões de termos gerais:

a) $u_{n}=\left( \dfrac{a}{1+\left\vert a\right\vert }\right) ^{n},\qquad\qquad\qquad$ b) $u_{n}=\sqrt[n]{\dfrac{\left( 3n\right) !}{\left( n!\right) ^{3}}}$ ,

em que $a$ é um número real.

Do meu blog, post de hoje.
ateixeira Says:
December 29, 2009 at 10:43 pm

a)

Firs let us determine the modulus of $\dfrac{a}{1+|a|}$

$\left | \dfrac{a}{1+|a|} \right | = \dfrac{|a|}{|1+|a||}=\dfrac{|a|}{1+|a|}$ since $1+|a|>0|$

Now $|a|< 1+|a|$ and so $\dfrac{|a|}{1+|a|}<1$ . Since the modulus of the expression is less than 1 we know that $\lim \left ( \dfrac{a}{1+|a|}\right )^n=0$

b) I'm lazy now but I'll write the solution tomorrow. Anyway the answer is 27. I did the calculations mostly in my head so maybe I'm getting the number wrong but I'm pretty sure that the limit is finite.
Américo Tavares Says:
December 29, 2009 at 11:12 pm

Hi!

In case you have any doubts you can always see the hint I wrote.

Bests,

Américo
ateixeira Says:
January 1, 2010 at 4:19 pm

Hi again!
Both problems are solved in my other blog.

Have a nice year!
Américo Tavares Says:
January 5, 2010 at 12:20 am

Olá!

a) está perfeito!

b) está certo. Há um caminho mais simples, o da sugestão. Mas gosto do seu método, que evita ter de conhecer a proposição para a qual a minha sugestão aponta. Se quizer pode colocar a sua resolução no meu blogue. Em alternativa, se me autorizar, coloco-a eu lá.

Obrigado

Américo
ateixeira Says:
January 5, 2010 at 1:29 pm

Pode colocar por mim e se quiser pode também comentar as minhas propostas de resolução.
Américo Tavares Says:
January 5, 2010 at 2:58 pm

Obrigado.