## Real Analysis Exercises III

1.

a) Calculate ${ \displaystyle \sum_{k=p}^{m}(u_{k+1}-u_k)}$ and ${\displaystyle\sum_{k=p}^{m}(u_k - u_{k+1}}$

${\displaystyle \sum_{k=p}^{m}(u_{k+1}-u_k)=u_{p+1}-u_{p}+u_{p+2}-u_{p+1}+\ldots +u_{m+1}-u_{m}}$

As we can see the first term cancels out with the fourth, the third with the sixth, and so on and all we are left with is the second and second last terms:

${\displaystyle \sum_{k=p}^{m}(u_{k+1}-u_k) = u_{m+1}-u_p}$

{\begin{aligned} \displaystyle \sum_{k=p}^{m}(u_k - u_{k+1})&= - \sum_{k=p}^{m}(u_{k+1}-u_k)\\ &= - (u_{m+1}-u_p)\\ &= u_p-u_{m+1} \end{aligned}}

b) Calculate ${\displaystyle \lim \sum_{k=1}^n\dfrac{1}{k(k+1)}}$ using the previous result.

${\displaystyle \lim \sum_{k=1}^n\dfrac{1}{k(k+1)}= \lim \sum_{k=1}^n \left( \frac{1}{k}-\frac{1}{k+1} \right) }$

Defining ${u_k=1/k}$ the previous sum can be written as

{\begin{aligned} \displaystyle \lim \sum_{k=1}^n \left( u_k-u_{k+1} \right)&=\lim (u_1 - u_{n+1})\\ &= \lim \left(1-\frac{1}{n+1}\right)\\ &=1 \end{aligned}}

This last result apparently has a funny story. Mengoli was the first one to calculate ${\displaystyle \lim \sum_{k=1}^n\dfrac{1}{k(k+1)}=1}$.

At the time this happened people did research in mathematics (I’m using this term rather abusively) in a somewhat different vein. They didn’t rush to print what they found like today.

Many times people held out their results for years while tormenting their rivals about what they found.

This is exactly what Mengoli did. In the times he was around the theory of series wasn’t much developed, thus this result, that we can calculate without being particularly brilliant in Mathematics, was something to take note of.

So, he wrote some letters to people saying that ${\displaystyle \lim \sum_{k=1}^n\dfrac{1}{k(k+1)}=1}$, but not how he concluded that.

The other mathematicians he sent the result too didn’t know about his methods and all they could do was to add numbers up explicitly and the only thing they could see was that even though they could sum more and more terms the result was always less than ${1}$ and was got nearer and nearer to ${1}$.

Of course this didn’t prove nothing since summing up a billion terms isn’t the same as summing an infinite number of terms and everyone but Mengoli was dumbfounded with that surprising result.

c) Calculate ${\displaystyle \sum_{k=0}^{n-1}(2k+1) }$

In this exercise what we are calculating is the sum of ${n}$ consecutive odd numbers. This result was already known to the ancient Greeks and the result wasn’t nothing short to astounding to them.

But enough with the talk already:

{\begin{aligned} \displaystyle \sum_{k=0}^{n-1}(2k+1)&=\sum_{k=0}^{n-1}\left[ (k+1)^2-k^2\right]\\ &= \sum_{k=0}^{n-1}(u_{k+1}-u_k) \end{aligned}}

With ${u_k=k^2}$

Using the now familiar formula

{\begin{aligned} \displaystyle \sum_{k=0}^{n-1}(2k+1) &= (n-1+1)^2-0^2\\ &= n^2 \end{aligned}}

An astounding result indeed!

Just look to ${\displaystyle \sum_{k=0}^{n-1}(2k+1)=n^2}$, interpret the result and try not to be as surprised as the ancient Greeks were.

2.

a) Using 1.a) and ${a^k=a^k\dfrac{a-1}{a-1}\quad (a \neq 1)}$ calculate ${\displaystyle \sum_{k=0}^{n-1} a^k }$

{\begin{aligned} \displaystyle \sum_{k=0}^{n-1} a^k &= \displaystyle\sum_{k=0}^{n-1} \left[ a^k\frac{a-1}{a-1}\right]\\ &= \displaystyle \frac{1}{a-1}\sum_{k=0}^{n-1}\left( a^{k+1}-a^k\right)\\ &= \displaystyle\frac{1}{a-1}(a^n-1)\\ &= \displaystyle\frac{a^n-1}{a-1} \end{aligned}}

b) Using a) establish the Bernoulli inequality ${a^n-1 \geq n(a-1)}$ if ${a > 0}$ and ${n \in \mathbb{Z}^+}$

If ${a=1}$ it is ${1-1=n(1-1) \Rightarrow 0=0}$ which is trivially true.

If ${n=1}$ it is ${a-1=a-1}$ which is trivially true.

For ${n \geq 2 }$ and ${a>1}$ it is:

{\begin{aligned} \displaystyle \sum_{k=0}^{n-1}a^k&= 1+a+a^2+\ldots+a^{n-1}\\ &> 1+1+\ldots+1\\ &= n \end{aligned}}

Thus

{\begin{aligned} \dfrac{a^n-1}{a-1} &> n \\ a^n-1 &> n(a-1) \end{aligned}}

Since ${a > 1}$

Finally if ${0 < a <1 }$ it is

{\begin{aligned} \displaystyle \sum_{k=0}^{n-1}a^k&= 1+a+a^2+\ldots+a^{n-1}\\ &< 1+1+\ldots+1\\ &= n \end{aligned}}

Thus

{\begin{aligned} \dfrac{a^n-1}{a-1} & < n \\ a^n - 1 & > n(a-1) \end{aligned}}

Since ${a < 1}$

c) Use b) to calculate ${\lim a^n}$ if ${a > 1}$ and then conclude that ${\lim a^n=0}$ if ${|a| < 1}$.

By b) it is

{\begin{aligned} a^n &> n(a-1)+1 \\ \lim a^n &\geq \lim \left( n(a-1)+1 \right)= +\infty \end{aligned}}

Hence ${\lim a^n = +\infty \quad (a>1)}$

For the second part of the exercise we will calculate instead ${\lim |a^n|}$ since that we know that ${ u_n \rightarrow 0 \Leftrightarrow |u_n| \rightarrow 0}$

Let us make a change of variable ${t=1/a}$. Thus ${|a|=|1/t|}$ and

{\begin{aligned} \lim |a^n| &= \lim |1/t|^n\\ &= \dfrac{1}{\lim |t|^n}\\ &= \dfrac{1}{+\infty}\\ &=0 \end{aligned}}

3. Consider the sequences ${u_n=\left( 1+\dfrac{1}{n} \right)^n }$ and ${v_n=\left( 1+\dfrac{1}{n} \right)^{n+1}}$

a) Calculate ${\dfrac{v_n}{v_{n+1}}}$ and ${\dfrac{u_{n+1}}{u_n}}$. Then use Bernoulli’s inequality to show that ${v_n}$ is strictly decreasing and that ${u_n}$ is strictly increasing.

{\begin{aligned} \dfrac{v_n}{v_{n+1}} &= \dfrac{\left( 1+1/n \right)^{n+1}}{\left(1+1/(n+1)\right)^{n+2}}\\ &=\dfrac{\left(\dfrac{n+1}{n}\right)^{n+1}}{\left( \dfrac{n+2}{n+1} \right)^{n+2}}\\ &= \dfrac{n}{n+1}\dfrac{\left(\dfrac{n+1}{n}\right)^{n+2}}{\left( \dfrac{n+2}{n+1} \right)^{n+2}}\\ &=\dfrac{n}{n+1}\left( \dfrac{(n+1)^2}{n(n+2)} \right)^{n+2}\\ &= \dfrac{n}{n+1}\left( \dfrac{n^2+2n+1}{n(n+2)} \right)^{n+2}\\ &=\dfrac{n}{n+1}\left( \dfrac{n(n+2)+1}{n(n+2)} \right)^{n+2}\\ &= \dfrac{n}{n+1}\left( 1+\dfrac{1}{n(n+2)} \right)^{n+2} \end{aligned}}

After having calculated ${v_n/v_{n+1}}$ we can use Bernoulli’s inequality, with ${a=1+\dfrac{1}{n(n+2)}}$ , to conclude that ${v_n}$ is strictly decreasing.

{\begin{aligned} \dfrac{n}{n+1}\left( 1+\dfrac{1}{n(n+2)} \right)^{n+2} &> \dfrac{n}{n+1}\left(1 + \dfrac{n+2}{n(n+2)} \right)\\ &= \dfrac{n}{n+1}(1+1/n)\\ &= \dfrac{n}{n+1}\dfrac{n+1}{n}\\ &= 1 \end{aligned}}

Thus ${v_n}$ is strictly decreasing.

With a similar technique we can prove that

${ \displaystyle u_{n+1}/u_n=\dfrac{n+1}{n}\left( 1- \dfrac{1}{(n+1)^2}\right)^{n+1}}$

After that by using Bernoulli’s inequality like in the previous example one can show that ${u_{n+1}/u_n>1}$ and thus ${u_n}$ is strictly increasing.

c) Using a) and b) and ${\lim u_n = e}$ prove the following inequalities ${(1+1/n)^n < e <(1+n)^{n+1}}$.

{\begin{aligned} \lim v_n&= \lim(1+1/n)^n(1+1/n)\\ &= e\times 1\\ &= e \end{aligned}}

We already know that ${v_n}$ is decreasing so it is ${v_n<(1+1/n)^{n+1}}$

On the other hand ${u_n}$ is increasing and ${\lim u_n=e}$ so ${(1+1/n)^n.

Hence ${(1+1/n)^n

d) Use c) to prove that ${ \displaystyle \frac{1}{n+1}<\log (n+1)-\log n <\frac{1}{n}}$

{ \begin{aligned} (1+1/n)^n &< e \\ n \log \left( \dfrac{n+1}{n} \right) &< 1 \\ \log(n+1) - \log n &< \dfrac{1}{n} \end{aligned} }

And now for the second part of the inequality:

{ \begin{aligned} e &< \left(1+\dfrac{1}{n}\right)^{n+1} \\ 1 &< (n+1)\log \left(\dfrac{n+1}{n}\right) \\ \dfrac{1}{n+1} &< \log (n+1) -\log n \end{aligned}}

In conclusion it is ${ \dfrac{1}{n+1}<\log (n+1)- \log n < \dfrac{1}{n} }$

4.

a) Using 3d) show that

${ \displaystyle 1+\log k < (k+1)\log (k+1)-k\log k < 1+ \log(k+1) }$

From

{ \begin{aligned} \dfrac{1}{k+1} &< \log (k+1) - \log k \\ 1 &< (k+1)\log(k+1) - (k+1)\log k \\ 1+ \log k &< (k+1)\log(k+1)-k \log k \end{aligned}}

With a similar reasoning we can also prove that ${(k+1)\log(k+1)-l\log k < 1+ \log(k+1)}$.

Thus it is ${1+\log k < (k+1)\log(k+1)-k\log k < 1+ \log(k+1)}$

b) Sum the previous inequalities between ${1 \leq k \leq n-1}$.

{\begin{aligned} \displaystyle \sum_{k=1}^{n-1}(1+ \log k) &< \sum_{k=1}^{n-1} ((k+1)\log(k+1)-k \log k)\\ &< \displaystyle \sum_{k=1}^{n-1}(1+\log(k+1)) \end{aligned}}

Now

{\begin{aligned} \displaystyle \sum_{k=1}^{n-1} (1+ \log k) &= \sum_{k=1}^{n-1}1+\sum_{k=1}^{n-1}\log k\\ &= n-1 +\sum_{k=1}^{n-1}\log k \end{aligned}}

And

{\begin{aligned} \displaystyle \sum_{k=1}^{n-1}\log k &= \log 1 + \log2 +\ldots+\log(n-1)\\ &=\log((n-1)!) \end{aligned}}

It also is

{\begin{aligned} \displaystyle \sum_{k=1}^{n-1}((k+1)\log(k+1) - k\log k)&= m\log n -\log 1\\ &=n\log n \end{aligned}}

And ${\displaystyle \sum_{k=1}^{n-1}(1+\log(k+1))=n-1+\log n!}$

Thus it is ${n-1+\log(n-1)! < n\log n < n-1 \log n!}$

c) Conclude the following inequalities ${ n \log n -n +1 < \log n! < n \log n -n+1+\log n}$ and establish Stirling's approximation ${ \displaystyle \log n! = n\log n -n +r_n}$ with ${e < C_n < en}$

{ \begin{aligned} n-1 + \log (n-1)! &< n\log n \\ \log (n-1)! &< n\log n -n+1 \\ \log n! &< n\log n -n +1+\log n \end{aligned}}

On the other hand

{\begin{aligned} n\log n &< n-1 + \log n! \\ n\log n -n +1 &< \log n! \end{aligned} }

Thus

{\begin{aligned} n\log n -n +1 &< \log n! \\ &< n\log n -n +1 +\log n \end{aligned}}

And from this follows ${1 < \log n! -n\log n+n < 1+\log n}$

Defining ${r_n=\log n! -n\log n+n}$ it is ${\log n! = n\log n-n+r_n}$ with ${1 < r_n < 1+\log n}$

5.

Show that ${\log \left(1+\dfrac{1}{n}\right)\sim \dfrac{1}{n}}$ and that ${\log \left(1+\dfrac{1}{n^2}\right)\sim \dfrac{1}{n^2}}$

We know that

{ \begin{aligned} \dfrac{1}{n+1} &< \log(n+1)-\log n < \dfrac{1}{n} \\ \dfrac{1}{n+1} &< \log\left( \dfrac{n+1}{n}\right) < \dfrac{1}{n} \\ \dfrac{1}{n+1} &< \log\left( 1+\dfrac{1}{n}\right) <\dfrac{1}{n} \\ \dfrac{1/(n+1)}{1/n} &< \dfrac{\log (1+1/n)}{1/n}<1 \\ \lim \dfrac{n}{n+1} &\leq \lim \dfrac{\log (1+1/n)}{1/n} \leq \lim 1 \\ 1 &\leq \lim \dfrac{\log (1+1/n)}{1/n} \leq 1 \end{aligned}}

Thus ${\lim \dfrac{\log (1+1/n)}{1/n}=1}$ and this equivalent to saying that ${\log \left(1+\dfrac{1}{n}\right)\sim \dfrac{1}{n}}$

Let ${u_n = \dfrac{\log (1+1/n)}{1/n}}$. In this case it is ${\dfrac{\log (1+1/n^2)}{1/n^2}=u_{n^2}}$. Since ${u_{n^2}}$ is a subsequence of ${u_n}$ we know that ${\lim u_{n^2}= \lim u_n}$ and so it also is ${\log \left(1+\dfrac{1}{n^2}\right)\sim \dfrac{1}{n^2}}$.

6. Show that ${u_n \sim v_n}$ and ${v_n \sim w_n \Rightarrow u_n \sim w_n }$

By hypothesis it is ${u_n=h_n v_n}$, ${v_n=t_n w_n}$ with ${h_n,t_n \rightarrow 1}$.

Substituting the second equality in the first we obtain ${u_n = h_n t_n w_n}$.

Let ${s_n = h_n t_n}$ and we write ${u_n =s_n w_n }$ with ${\lim s_n = \lim h_n \lim t_n =1\times 1=1}$.

Thus ${u_n \sim w_n}$

7. Let ${u_n = O\left(1/n\right)}$ and ${v_n = O (1/ \sqrt{n})}$. Show that ${u_n v_n = o ( 1/n^{4/3})}$.

${u_n = h_n 1/n}$ and ${v_n = t_n 1/ \sqrt{n}}$ with ${h_n}$ and ${t_n}$ bounded sequences. Now

{\begin{aligned} u_n v_n &= \dfrac{h_n}{n} \dfrac{t_n}{\sqrt{n}}\\ &= \dfrac{h_n t_n}{n^{3/2}}\\ &=\dfrac{h_n t_n}{n^{1/6}}\dfrac{1}{n^{4/3}} \end{aligned}}

Let ${s_n = \dfrac{h_n t_n}{n^{1/6}}}$ it is ${\lim s_n = \lim \dfrac{h_n t_n}{n^{1/6}} = 0}$ since ${h_n t_n}$ is bounded.

Thus ${u_n v_n = o (1/n^{4/3})}$

8. Using Stirling’s approximation show that ${\log n! = n\log n -n + O(\log n)}$

We know that it is ${\log n! = n\log n -n + +r_n}$ with ${ 1< r_n < 1+\log n}$. Thus

{\begin{aligned} 0 &<\dfrac{1}{\log n}\\ &< \dfrac{r_n}{\log n}\\ &< \dfrac{1}{\log n} +1\\ &\leq \dfrac{1}{\log 2}+1 \end{aligned}}

Where we used the fact that ${ \dfrac{1}{\log n}+1}$ is decreasing function.

Thus ${\dfrac{r_n}{\log n}}$ is bounded and so ${r_n=O(\log n)}$ as desired.

### 6 Responses to “Real Analysis Exercises III”

1. Prove

$\displaystyle\sum_{n=1}^{\infty }\sqrt{\arcsin \left( \sqrt{\dfrac{n^{2}}{1+n^{2}}}\right)+\arctan \left( n-1\right) -2\sqrt{\arctan\left( n\right) \cdot\arcsin\left( \sqrt{\dfrac{\left( n-1\right) ^{2}}{1+\left( n-1\right) ^{2}}}\right)}}$ $=\sqrt{\dfrac{\pi }{2}}$.

2. Em português:

Muitas vezes uma série é telescópica, mas nem sempre é fácil reconhecer esse facto.

Exercício: Sejam $a_n$ e $b_n$ respectivamente
$a_n=\arcsin (b_n)+\arctan (n-1)-2\sqrt{\arctan (n)\cdot\arcsin (b_{n-1})}$
e
$b_n=\dfrac{n}{\sqrt{1+n^2}}$.
Mostre que $\displaystyle\sum_{n=1}^{\infty }\sqrt{a_n}=\sqrt{\dfrac{\pi }{2}}$.
Resolução: a única dificuldade é mostrar que a série é telescópica. Vamos aproveitar a identidade trigonométrica provada neste problema:
$\arctan\left( \dfrac{u}{\sqrt{1-u^{2}}}\right) =\arcsin u$.
Dela obtém-se
$\arctan \left( x\right) =\arcsin \left( \sqrt{\dfrac{x^{2}}{1+x^{2}}}\right)$,
fazendo a substituição $u=\sqrt{\dfrac{x^{2}}{1+x^{2}}}$ ( equivalente a $x=\dfrac{u}{\sqrt{1-u^{2}}}$). Assim, temos
$\arcsin (b_n)=\arcsin\left(\dfrac{n}{\sqrt{1+n^2}}\right) =\arctan \left( n\right)$
$\arcsin (b_{n-1})=\arcsin\left(\dfrac{n-1}{\sqrt{1+{n-1}^2}}\right) =\arctan \left( {n-1}\right)$
donde
$a_n=\arctan \left( n\right) +\arctan (n-1)-2\sqrt{\arctan (n)\cdot\arctan \left( {n-1}\right) }$.
Pondo $u_{n}=\arctan \left( n\right)$ e atendendo à relação algébrica
$\sqrt{u_{n}+u_{n-1}-2\sqrt{u_{n}u_{n-1}}}=\sqrt{u_{n}}-\sqrt{u_{n-1}}$
chegamos efectivamente à série telescópica
$\displaystyle\sum_{n=1}^{\infty }\sqrt{a_{n}}=\displaystyle\sum_{n=1}^{\infty }\sqrt{u_{n}+u_{n-1}-2\sqrt{u_{n}u_{n-1}}}$
$=\displaystyle\sum_{n=1}^{\infty }\sqrt{u_{n}}-\sqrt{u_{n-1}}=\underset{N\rightarrow \infty }{\lim }\sqrt{u_{N}}-\sqrt{u_{0}}$
$=\underset{N\rightarrow \infty }{\lim }\sqrt{\arctan \left( N\right) }-\sqrt{\arctan \left( 0\right) }=\underset{N\rightarrow \infty }{\lim }\sqrt{\arctan \left( N\right) }=\sqrt{\dfrac{\pi }{2}}\blacktriangleleft$
Repetindo, muitas vezes uma série é telescópica, mas nem sempre é fácil reconhecer esse facto. Com este exemplo pretendi ilustrar uma situação de dificuldade intermédia, avaliação que é claramente subjectiva porque depende muito de resultados anteriores que se conhecem ou não: neste caso, uma identidade trigonométrica.

http://problemasteoremas.wordpress.com/2009/06/09/serie-telescopica-irracional-e-trigonometrica/

3. Eu tinha um pressentimento que a soma era telescópica, mas não consegui demonstrá-lo por mim mesmo (acho que se tivesse recorrido aos meus apontamentos o teria resolvido, mas sinceramente foi preguiça – e também porque gosto de resolver as coisas por mim).

E já viu as minhas outras resoluções?

4. Sim, estão ambas certas. Deixei lá um comentário.

Américo

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