# Thread: A Relation About Bernoulli Numbers and Theorem of Arithemtic

1. ## A Relation About Bernoulli Numbers and Theorem of Arithemtic

Hi,

As I posted in to number theory forum, I noticed that

$\displaystyle N^{s+1}\equiv N \quad Denominator[B_s]$

where $\displaystyle B_s$ is the $\displaystyle s-th$ Bernoulli Number

I think that i have also a correct proof.

Here I also have the opportunity to thank this forum not olny for the help prividing me, but also for giving me the chance of expressing my concerns in mathematics.

2. Originally Posted by gdmath
Hi,

As I posted in to number theory forum, I noticed that

$\displaystyle N^{s+1}\equiv N \quad Denominator[B_s]$

where $\displaystyle B_s$ is the $\displaystyle s-th$ Bernoulli Number

I think that i have also a correct proof.

Here I also have the opportunity to thank this forum not olny for the help prividing me, but also for giving me the chance of expressing my concerns in mathematics.
I haven't read all the way through this yet, but I have read through your proof of your test for series convergence.

In the test of convergence your statement of the test and notation needs to be improved.

It could read something like:

Test of Convergence

The series $\displaystyle \sum_{i=1}^{\infty} a_n$ converges if:

$\displaystyle \lim_{n\to \infty} \frac{a_n a_{n-1}}{a_n-a_{n-1}}=0$

Also I cannot follow your argument here:

Please inprove the notation and if possible clarify the reasoning.

(I will leave you post here for two weeks for you to prot a revision after that I will removing it as the standard of presentation is at present not acceptable)

CB

3. About the test of convergence i based on ratio test proof as i found it here :
http://calculus7.com/sitebuildercont...fratiotest.rtf

At the point of

$\displaystyle \sum_{n=1}^{\infty}\frac{f(i)}{p-n\cdot p^{n-1}\cdot f(i)}$

I mean that i have a new series with close form:

$\displaystyle a_n=\frac{f(i)}{p-n\cdot p^{n-1}\cdot f(i)}$

and i prove that this series is convergent throught ratio test.

Since $\displaystyle a_n>f(i+n)$, then throught the comparison test if $\displaystyle a_n$converges then also $\displaystyle f(i+n)$ converges.

However my doubt is at the beggining, when i solve only the right inequality:

$\displaystyle -r<\frac{f(j)\cdot f(j+1)}{f(j+1)-f(j)}<r$

Please tell me if there is a fault point (because i do not concern about the left inequality).

Anyway the convergence test might not belong in this paper (beyond the impovments that proof needs).

I include it in the corrected paper only if we are 100% certain that proof is correct and proper typed.

Thank you a lot.

4. ## revisited

I have made a revision of the document.

Should i wait this thread to be deleted (and post new one) or leave it here?