## Real Analysis Exercises III

1.

a) Calculate and

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:

b) Calculate using the previous result.

Defining the previous sum can be written as

This last result apparently has a funny story. Mengoli was the first one to calculate .

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 , 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 and was got nearer and nearer to .

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

In this exercise what we are calculating is the sum of 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:

With

Using the now familiar formula

An astounding result indeed!

Just look to , interpret the result and try not to be as surprised as the ancient Greeks were.

2.

a) Using 1.a) and calculate

b) Using a) establish the Bernoulli inequality if and

If it is which is trivially true.

If it is which is trivially true.

For and it is:

Thus

Since

Finally if it is

Thus

Since

c) Use b) to calculate if and then conclude that if .

By b) it is

Hence

For the second part of the exercise we will calculate instead since that we know that

Let us make a change of variable . Thus and

3. Consider the sequences and

a) Calculate and . Then use Bernoulli’s inequality to show that is strictly decreasing and that is strictly increasing.

After having calculated we can use Bernoulli’s inequality, with , to conclude that is strictly decreasing.

Thus is strictly decreasing.

With a similar technique we can prove that

After that by using Bernoulli’s inequality like in the previous example one can show that and thus is strictly increasing.

c) Using a) and b) and prove the following inequalities .

We already know that is decreasing so it is

On the other hand is increasing and so .

Hence

d) Use c) to prove that

And now for the second part of the inequality:

In conclusion it is

4.

a) Using 3d) show that

From

With a similar reasoning we can also prove that .

Thus it is

b) Sum the previous inequalities between .

Now

And

It also is

And

Thus it is

c) Conclude the following inequalities and establish Stirling's approximation with

On the other hand

Thus

And from this follows

Defining it is with

5.

Show that and that

We know that

Thus and this equivalent to saying that

Let . In this case it is . Since is a subsequence of we know that and so it also is .

6. Show that and

By hypothesis it is , with .

Substituting the second equality in the first we obtain .

Let and we write with .

Thus

7. Let and . Show that .

and with and bounded sequences. Now

Let it is since is bounded.

Thus

8. Using Stirling’s approximation show that

We know that it is with . Thus

Where we used the fact that is decreasing function.

Thus is bounded and so as desired.

December 30, 2009 at 8:27 am

Prove

.

January 4, 2010 at 6:24 pm

Em português:

Muitas vezes uma série é telescópica, mas nem sempre é fácil reconhecer esse facto.Exercício: Sejam e respectivamentee

.

Mostre que .

Resolução:a única dificuldade é mostrar que a série é telescópica. Vamos aproveitar a identidade trigonométrica provada neste problema:.

Dela obtém-se

,

fazendo a substituição ( equivalente a ). Assim, temos

donde

.

Pondo e atendendo à relação algébrica

chegamos efectivamente à série telescópica

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.Publicado originalmente em

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

January 4, 2010 at 10:57 pm

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?

January 5, 2010 at 12:24 am

Sim, estão ambas certas. Deixei lá um comentário.

Américo

January 5, 2010 at 11:50 am

Em

https://climbingthemountain.wordpress.com/2009/02/20/real-analysis-exercises-ii/#comment-91

February 15, 2014 at 7:00 pm

[…] denote the nearest integer function and using Bernoulli’s Inequality () it is […]