The proposed 'chellenge' is retired because it has been badly impostated!...

I beg your pardon for the time You have wasted!...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Printable View

- Nov 12th 2009, 12:25 AMchisigmaAbout Euler's constant (4)...
The proposed 'chellenge' is retired because it has been badly impostated!...

I beg your pardon for the time You have wasted!...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Nov 12th 2009, 06:58 AMDrexel28
- Nov 13th 2009, 01:57 AMchisigma
In order to avoid misunderstanding a preliminary question must be answered...

In the following previous thread...

http://www.mathhelpforum.com/math-help/number-theory/112730-question-about-eulers-phi-function.html

... starting from the basic explicit formula for $\displaystyle \varphi(n)$...

$\displaystyle \varphi(n)= n\cdot \prod_{p|n} (1-\frac{1}{p})$ (1)

... first we derived ...

$\displaystyle \frac{\varphi(n)}{n} = \prod_{p|n} (1-\frac{1}{p})$ (2)

... and from which we demonstrated that...

$\displaystyle \frac{\varphi(n)}{n} \ge \prod _{p \le n} (1-\frac{1}{p}) $ (3)

Now from (3) I realize that is...

$\displaystyle \inf \{\frac{\varphi(n)}{n}\} = \prod _{p \le n} (1-\frac{1}{p}) $ (4)

The question: is all that correct?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Nov 14th 2009, 09:38 PMchisigma

In order to avoid any possible confusion I recast the previous question in more general terms...

The possibilities that the expression $\displaystyle \liminf_{n \rightarrow \infty} x (n)$ does have any sense are two...

a) given a sequence $\displaystyle x_{n}$ at it is asociated one and only one $\displaystyle f(n) = inf\{x(n)\}$ and we are interested to the $\displaystyle \lim_{n \rightarrow \infty} f(n)$. In this case we need to have an acceptable definition of $\displaystyle f(n)$ ...

b) given a sequence $\displaystyle x_{n}$ it is said to have a 'lower bound' $\displaystyle a$ if it doesn't exist any $\displaystyle n_{0} \in \mathbb{N}$ for which is $\displaystyle x(n_{0}) < a$ and , given a $\displaystyle \varepsilon >0$, it exist at least one $\displaystyle n_{0} \in \mathbb{N}$, for which is $\displaystyle a \le x(n_{0}) \le a +\varepsilon$. In this case however the expression $\displaystyle \liminf_{n \rightarrow \infty}$ is a nonsense...

The question : which is true?... a) or b)?... or something else?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Nov 16th 2009, 07:37 AMDrexel28
I'm not exactly sure what the confusion here is. If we let $\displaystyle S$ be the set of all subsequential limits of $\displaystyle \left\{\frac{\varphi(n)\ln\ln n}{n}\right\}_{n=1}^{\infty}$ (i.e. $\displaystyle \lim_{n\to\infty}\frac{\varphi(a_n)\ln \ln a_n}{a_n}$ for any, increasing $\displaystyle a_n$) then $\displaystyle \liminf_{n\to\infty}\frac{\varphi(n)\ln \ln n}{n}=\inf\text{ }S$