# Math Help - About Euler's constant (4)...

1. ## About 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

$\chi$ $\sigma$

2. Originally Posted by chisigma
The proposed 'chellenge' retired becuase it has been badly impostated!...

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

Kind regards

$\chi$ $\sigma$
Prove that $\liminf_{n\to\infty}\frac{\varphi(n)\ln\ln(n)}{n}= e^{-\gamma}$, whre $\varphi$ is Euler's totient function. Try that on for size.

3. Originally Posted by Drexel28
Prove that $\liminf_{n\to\infty}\frac{\varphi(n)\ln\ln(n)}{n}= e^{-\gamma}$, whre $\varphi$ is Euler's totient function....
In order to avoid misunderstanding a preliminary question must be answered...

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

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

... first we derived ...

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

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

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

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

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

The question: is all that correct?...

Kind regards

$\chi$ $\sigma$

4. Originally Posted by Drexel28
Prove that $\liminf_{n\to\infty}\frac{\varphi(n)\ln\ln(n)}{n}= e^{-\gamma}$, whre $\varphi$ is Euler's totient function. Try that on for size.

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

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

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

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

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

Kind regards

$\chi$ $\sigma$

5. Originally Posted by chisigma
In order to avoid any possible confusion I recast the previous question in more general terms...

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

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

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

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

Kind regards

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