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Bibubo
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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.
Bibubo said: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.
mathbalarka said:Ideally, a proof would probably involve the Euler product instead of the Dirichlet series
$$\frac{\zeta(s)}{\zeta(2s)} = \prod_p \left ( 1 + \frac1{p^s} \right )$$
chisigma said: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 said: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 said: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
$\chi$ $\sigma$
The Zeta Function, denoted as ζ(s), is a mathematical function that is used to study the distribution of prime numbers. It is related to inequality as it can be used to prove the Riemann Hypothesis, which states that all non-trivial zeros of the Zeta Function lie on the critical line with a real part of 1/2.
The Riemann Hypothesis is one of the most famous unsolved problems in mathematics. It states that all non-trivial zeros of the Zeta Function lie on the critical line with a real part of 1/2. It is important because it has many implications in number theory, and its proof would have a significant impact on the field of mathematics.
The Zeta Function has a close connection to prime numbers through the Euler product formula. This formula expresses the Zeta Function as an infinite product involving all prime numbers, giving insight into the distribution of primes.
Some common techniques used to study inequalities involving the Zeta Function include analytic methods, probabilistic methods, and algebraic methods. These techniques involve manipulating the properties of the Zeta Function to prove various inequalities.
Inequalities involving the Zeta Function have various applications in number theory, analysis, and geometry. They have been used to study the distribution of prime numbers, estimate the size of certain sets, and investigate the behavior of certain functions.