MHB Inequality involving Zeta Function

[Bibubo]

Prove that for $r>2$ we have $$\frac{\zeta\left(r\right)}{\zeta\left(2r\right)}<\left(1+\frac{1}{2^{r}}\right)\frac{\left(1+3^{r}\right)^{2}}{1+3^{2r}}.$$ I've tried to write Zeta as Euler product but I haven't solve it.

 

[chisigma]

Prove that for $r>2$ we have $$\frac{\zeta\left(r\right)}{\zeta\left(2r\right)}<\left(1+\frac{1}{2^{r}}\right)\frac{\left(1+3^{r}\right)^{2}}{1+3^{2r}}.$$ I've tried to write Zeta as Euler product but I haven't solve it.

There is an explicit expression of the function in term of Dirichlet series...

$\displaystyle \frac{\zeta (r)}{\zeta (2\ r)} = \sum_{n=1}^{\infty} \frac{\mu^{2} (n)} {n^{r}}\ (1)$

... where $\mu(n)$ is the Moebious function. May be that the (1) is the right way but it requires some work...

Kind regards

$\chi$ $\sigma$

 

[mathbalarka]

Ideally, a proof would probably involve the Euler product instead of the Dirichlet series

$$\frac{\zeta(s)}{\zeta(2s)} = \prod_p \left ( 1 + \frac1{p^s} \right )$$

 

[Bibubo]

Ideally, a proof would probably involve the Euler product instead of the Dirichlet series

$$\frac{\zeta(s)}{\zeta(2s)} = \prod_p \left ( 1 + \frac1{p^s} \right )$$

I've tried the Euler product way, but I haven't found a solution.

 

[Petek]

I've been working on this problem, but no solution. Where did it come from? I observed that, since $\zeta(2)= \frac{\pi^2}{6}$ and $\zeta(4)=\frac{\pi^4}{90}$, we have

$$\frac{\zeta(2)}{\zeta(4)}=\frac{15}{\pi^2}\approx 1.51981$$

and

$(1+\frac{1}{4})\frac{(1+9)^2}{1+81}\approx 1.52439$

so the inequality also holds for $r=2$.

 

[chisigma]

Writing the function as Euler product...

$\displaystyle \varphi(r)= \frac{\zeta(r)}{\zeta(2\ r)} = \prod_{p} (1 + \frac{1}{p^{r}}) = (1 + \frac{1}{2^{r}})\ (1 + \frac{1}{3^{r}})\ ... (1)$

... and taking the logarithm we obtain...

$\displaystyle \ln \frac{\varphi(r)}{1 + \frac{1}{2^{r}}} = \ln (1 + \frac{1}{3^{r}}) + ...\ (2)$

But is...

$\displaystyle \ln (1 + \frac{1}{3^{r}}) = \frac{1}{3^{r}} - \frac{1}{2\ 3^{2\ r}} + \frac{1}{3\ 3^{3\ r}} + ...\ (3)$

... and ...

$\displaystyle \ln \frac{(1 + \frac{1}{3^{r}})^{2}}{1 + \frac{1}{3^{2\ r}}} = \frac{2}{3^{r}} - \frac{1}{3^{2\ r}} + \frac{2}{3\ 3^{3\ r}} - ... - \frac{1}{3^{2\ r}} + \frac{1}{2\ 3^{4\ r}} - ... = \frac{2}{3^{r}} - \frac{2}{3^{2\ r}} + \frac{2}{3\ 3^{3\ r}} + ...\ (4)$

... so that comparing (3) and (4) we obtain that...

$\displaystyle \ln \frac{\varphi(r)}{1 + \frac{1}{2^{r}}} < \ln \frac{(1 + \frac{1}{3^{r}})^{2}}{1 + \frac{1}{3^{2\ r}}}\ (5)$

... what we intend to demonstrate...

Kind regards

$\chi$ $\sigma$

 

[Bibubo]

Writing the function as Euler product...

$\displaystyle \varphi(r)= \frac{\zeta(r)}{\zeta(2\ r)} = \prod_{p} (1 + \frac{1}{p^{r}}) = (1 + \frac{1}{2^{r}})\ (1 + \frac{1}{3^{r}})\ ... (1)$

... and taking the logarithm we obtain...

$\displaystyle \ln \frac{\varphi(r)}{1 + \frac{1}{2^{r}}} = \ln (1 + \frac{1}{3^{r}}) + ...\ (2)$

But is...

$\displaystyle \ln (1 + \frac{1}{3^{r}}) = \frac{1}{3^{r}} - \frac{1}{2\ 3^{2\ r}} + \frac{1}{3\ 3^{3\ r}} + ...\ (3)$

... and ...

$\displaystyle \ln \frac{(1 + \frac{1}{3^{r}})^{2}}{1 + \frac{1}{3^{2\ r}}} = \frac{2}{3^{r}} - \frac{1}{3^{2\ r}} + \frac{2}{3\ 3^{3\ r}} - ... - \frac{1}{3^{2\ r}} + \frac{1}{2\ 3^{4\ r}} - ... = \frac{2}{3^{r}} - \frac{2}{3^{2\ r}} + \frac{2}{3\ 3^{3\ r}} + ...\ (4)$

... so that comparing (3) and (4) we obtain that...

$\displaystyle \ln \frac{\varphi(r)}{1 + \frac{1}{2^{r}}} < \ln \frac{(1 + \frac{1}{3^{r}})^{2}}{1 + \frac{1}{3^{2\ r}}}\ (5)$

... what we intend to demonstrate...

Kind regards

$\chi$ $\sigma$

Sorry but I don't see the point. You have something like $$\log\left(\frac{\left(1+3^{r}\right)^{2}}{1+3^{2r}}\right)=2\left(\underset{n\, odd}{\sum}\frac{1}{n}\,\frac{1}{3^{rn}}-2\underset{n\, even,\,4\nmid n}{\sum}\frac{1}{n}\,\frac{1}{3^{2rn}}\right)$$ because if $4\mid n$ we have a cancellation. Why is this bigger than $$\underset{p\geq3}{\sum}\log\left(1+\frac{1}{p^{r}}\right)?$$

 

[chisigma]

Sorry but I don't see the point. You have something like $$\log\left(\frac{\left(1+3^{r}\right)^{2}}{1+3^{2r}}\right)=2\left(\underset{n\, odd}{\sum}\frac{1}{n}\,\frac{1}{3^{rn}}-2\underset{n\, even,\,4\nmid n}{\sum}\frac{1}{n}\,\frac{1}{3^{2rn}}\right)$$ because if $4\mid n$ we have a cancellation. Why is this bigger than $$\underset{p\geq3}{\sum}\log\left(1+\frac{1}{p^{r}}\right)?$$

I apologize for having understood the identity $\displaystyle \frac{(1 + 3^{r})^{2}}{1 + 3^{2\ r}}= \frac{(1 + \frac{1}{3^{r}})^{2}}{1 + \frac{1}{3^{2\ r}}}$, that permits to me to use the Taylor expansion $\displaystyle \ln (1 + \varepsilon) = \varepsilon - \frac{\varepsilon^{2}}{2} + \frac{\varepsilon^{3}}{3} - ...$...

Kind regards

$\chi$ $\sigma$

 

[Bibubo]

I apologize for having understood the identity $\displaystyle \frac{(1 + 3^{r})^{2}}{1 + 3^{2\ r}}= \frac{(1 + \frac{1}{3^{r}})^{2}}{1 + \frac{1}{3^{2\ r}}}$, that permits to me to use the Taylor expansion $\displaystyle \ln (1 + \varepsilon) = \varepsilon - \frac{\varepsilon^{2}}{2} + \frac{\varepsilon^{3}}{3} - ...$...

Kind regards

$\chi$ $\sigma$

You can write as you told. In fact you have $$\frac{\left(1+3^{r}\right)}{1+3^{2r}}^{2}=\frac{\left(3^{r}\left(1+\frac{1}{3^{r}}\right)\right)^{2}}{3^{2r}\left(1+\frac{1}{3^{2r}}\right)}=\frac{\left(1+\frac{1}{3^{r}}\right)^{2}}{1+\frac{1}{3^{2r}}}$$ but I haven't understood how your proof can show my inequality.