For the mads about analysis!

But where to begin? Well, let us attend to the equality containing the numbers of
Bernoulli, formula 1). Several demonstrations are possible, more
or less complex... The one below presented, has the merit to be fast (everything
is relative!) and is based on the first definition of the numbers of Bernoulli. it
shows besides by completing the following demonstration of the formula 2) the equivalence
of the two definitions of the numbers of Bernoulli...

Then just later, the original "demonstration"
(hem!) due to Euler for the sum of squares inverses. If rigour is not really the
first preoccupation, what an imagination and what a craftiness!

But that, it is for after the real proof...

1) We will be able to say indeed never enough
thank you to Mister Fourier, its theory is going to be still very useful for us...

Considering the polynomials of Bernoulli *B*_{2n}(t), let us take a function *f*_{2n} for an integer *n*, of period *1*, and that is equal to *B*_{2n} on *[0,1
[*. *F*_{2n} is obviously *C*^{1} by fragments because restriction of polynomes
in an interval of *R*. She verifies so the theorem of Dirichlet for *T* - periodic functions
and one can write it like a series of Fourier.

1.1 First small intermediate result:
1.2 Let us
calculate Fourier's coefficients now. Not to get bored pointlessly, let us
show first of all by recurrence on n that *B*_{2n} is even with regard to *x=1/2*...

*B*_{0}=1 by immediate calculation so no problems for the rank *1*...

Let us suppose for some integer *n* that *B*_{2n}(x+1/2)=B_{2n}(1/2-x)

Put *g*_{2n+2}(x)=B_{2n+2}(x+1/2)-B_{2n+2}(1/2-x). It is a polynôme so *g*_{2n+2} is *C*^{2} on *R*.

According to the hypothesis of recurrence...

So *g '*_{2n+2 }is constant now, *g'*_{2n+2}(0)=(2n+2)B_{2n+1}(0)-(2n+2)B_{2n+1}(0)=0 so *g'*_{2n+2}=0 so *g*_{2n+2} is constant now *g*_{2n+2}(0)=0 and finally, *B*_{2n+2} is very symmetric with regard to *x=1/2*, it is hypothesis
in the following rank so proposition is valid for every integer *n* and on *[0,1]*. *f*_{2n} is symmetric with regards to *x=1/2*.

So the associated Fourier coefficient is obtained with *b*_{k}(f_{2n})=

1.3 Let
us calculate on the other hand for every integer n not nil

*a*_{k}(B_{2n})=

*=* According to 1.1 then by immediate downward recurrence...

To there, everything is well! The rest is not much more complicated... We have indeed
to calculate *a*_{k}(B_{2}).

1.4 One
will be able to verify (because there is uniqueness) that *B*_{2}=X^{2}-X+1/6 is well the Bernoulli's polynomial
which is convenient for definitions. Then by simple
double integration by parts, one has and . And naturally, , so, finally,

*a*_{k}(B_{2})= .

1.5 One
found so according to 1.3 and 1.4

*a*_{k}(B_{2n})=

1.6 The development
of B_{2n }In series of Fourier
give us then for t in [0,1 [ :

and then specificly for t=0, one has:

where *Ber*_{2n}=B_{2n}(0)

And here is !

In view of the importance of the result, this demonstration is monstrously quick
with regard to others ! And by patching a little, also two very useful expressions
are obtained :

and

More generally, theses expressions are called Reynolds's
sums :

So, for *p=2n* odd, we obtain the above sum, and for *p* even, one has:

with En
the Euler's numbers that could be defined by :

or we
can use the Taylor's developpement :

. We have
in particular *E*_{2n-1}=0, E_{0}=1,
E_{2}=-1, E_{4}=5, E_{6}=-61, E_{8}=1385

These numbers bring in Euler's polynomials of which I did not regrettably find definition.
It is enough to apply the previous method to Euler's polynomials to find result.

Otherwise, with the first values of the numbers of Bernoulli and the first formula,
one obtains:

Among others...

1) Encore... Here is the so original proof,
of Euler himself, of the sum of invert squares... Convergence had been demonstrated
by Jacques Bernoulli
and Leibniz comes a cropper over,
let us see what Euler made. He found the limit, but justifications are a little doubtful,
you will see!

Basing itself on the theory of algebraic equations, Euler knew that the sum of invert
roots of *a*_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+1=0 worth *-a*_{1}. Considering then the development of It knows thanks to the periodicity
of sine that the right member is nil for *x=*^{2},4^{2},9^{2},16^{2}...

Euler says to himself whether the property is true for a finished polynôme it
must be for an infinite polynôme and he deducts from it Is . Very strong !

2) Check, let us return in a little more serious
things (mathematically!)... The demonstration of the general formula of the 2) (abbreviated,
let us not misuse the calculation, and any more, I feel a little tired!) is on Fourier's page. An application *z=1/2* gives :
.

But there is stronger! In one develops by exchanging sums (*z* inferior
to *1* and absolute convergence of sums)

By posing *t=2iZ* and by calculating cotan by means of its expression in exponential complex (*cos / sin*),
one obtains

*=*

According to the 2-nd definition of the numbers
of Bernoulli. Oh oh!

By identification of terms two for two (it is two series), one can then end

!

It is effectively the good result shown in 1) and
it is as well the original method that had used Euler. Here we are, *2* proofs for
the price for one! It implies also the equivalence of the definitions of the numbers
of Bernoulli (those of odd rank are indeed nil)

3) Ah, small break...

Well, let us move on the formula *3* now... Even there,
Euler's monstrous craftiness intervenes... This one uses indeed a very unusual development,
but o how effective, of *arctan* that is:

Arctan(t)=

For *t* positive... Euler used it with a formula of *arctan =20Arctan (1/7) +8Arctan (3/79)* to calculate *20* decimal of *Pi* in one hour (try so!). It is real that if one poses *t=1/7*, =0,02 ! Useful...
Skillful, Euler was certainly...

Let us see indeed the demonstration:

3.1 Let us put *x=* *x=*, *t* positive and *x* in *[0,1 [*.

One has t=. Let
us put y=Arctan(t)=Arctan

3.2 Let us try to simplify the writing... If
z=Arctan, one
obtains what gives sin^{2}z=x and so

y=Arcsin.

But one does not see there always very clear...
Let us continue so...

3.3 We
derive this expression in x and one obtains :

*2y'(x-x*^{2})+y(1-2x)=1

3.4 Let us
look there under the shape *y=*.
Inserted into the differential equation, our eyes popping out of their sockets discover *a*_{0}+x(3a_{1}-2a_{0})+=1+0.x+0.x^{2}+... That, by identification, gives us *a*_{0}=1, 3a_{1}-2a_{0}=0 so* a*_{1}=2/3, and* (2n+1)a*_{n}-2na_{n}-1=0 so :

By immediate recurrence on integer *n* (hyper classic
calculation!).

Mutually, one verifies that *y=* agrees how solution of the differential equation (not to forget, but do not count
on me to make it!).

3.5 The rest is coarse as say some... According
to the definition of *y*, we conclude for *t* positive real :

*Arctan(t)=*

But what can we make of it? Height of things! First
of all apply described Euler's formula at the beginning of this section... May pose
t=*1* and obtain the formula 3) or *t=1/2*, or *1/4*, or to use *Arctan*'s formulae everything is possible! As usual, the more *t* is small,
the more convergence is fast, because term in t in the sum is more and more small...

Me, I like *t=*,
because we obtain so *Arctan(t)=/6* :

That is very close to Schroeppel-Gosper's formula. Terrific!

4) Check,
another proof... Even if I appear to be somebody which abbreviates more and more
in the course of demonstrations, I can only give the tracks of the demonstration
of the formula 4)... According to *"Le
fascinant nombre Pi"* (see Ref.):

Is a> 1. For p prime number, *p*^{a}>1 so by limited development :
If these equalities for any prime number are written
and that their product is made, one obtains to the left :

And to the right then? The product of right members
is obtained, it is to say infinite sums. If one eliminates all the infinite products
which are going to form (infinite products of terms in *1/p*^{ka} give 0 would be it that because *0<1/p*^{ka}<1/2), it stays the sum all the terms of the shape , are prime numbers and *k*, *i*, *j* complete...
Does not it call back to you? Some theorem of arithmetic which says that quite complete
n is decomposable under the shape *p*^{k}q ^{i}r ^{j}... It is to
say as product of prime numbers each to some power... As all the prime numbers and
their powers are present here, one deducts from it that this infinite sum is worth
:

And for has peer, one has then according to the
formula 1) :

That is well the formula 4), if *Pi* is isolated!

These fabulous two equalities found by Euler will be reused in the page dedicated
to Cèsaro. One of the important consequences of this result is that
for *a=1*,
the sum of invert diverges, so the product (left member) is worth infinity, so there
is an infinity of prime numbers... Not bad, no? Here is that *Pi* makes a deafening
entry to the world of the arithmetic!