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I'm studying a book, which wants to prove that for some constants $C,\kappa$, we have $$\left|\frac{\zeta ''(s)\zeta (s) -2\zeta '(s)}{\zeta (s)^2}\right|\le C*|t|^\kappa~~~~,\text{for }s=\sigma+it,~|t|\ge 1, \sigma >1.$$

It has proved so far, that for every $m\in \mathbb{N}_0,$ we have constants $C_m$ with $$|\zeta^{(m)}(s)|\le C_m|t|~~~\text{, for }|t|\ge 1,\sigma>1$$ and for some $\delta>0$ we have: $$|\zeta(s)|\ge\delta|t|^{-4}~~~\text{, for }|t|\ge 1,\sigma>1.$$

It's obvious that the first inequality holds for e.g. $\kappa=9$. But my book says that the first inequality is true for $\kappa=5,$ too and I fail to see how we can prove this using these inequalities.

How can I prove the first inequality for $\kappa =5$? (is it even possible?)

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  • $\begingroup$ What book are you reading? $\endgroup$ – davidlowryduda Jun 19 '15 at 8:42
  • $\begingroup$ It's a german book by E. Freitag and R. Busam: "Funktionentheorie 1" (beginning of chapter 6 in the 4th edition of the book). The result is needed for the proof of the prime number theorem. The exact value of $\kappa$ is not relevant for the proof, but it's stated that the inequality holds for e.g. $\kappa=5$. As is I missed no obvious solution apparently, I guess it's just a mistake in the book as they stated that the first inequality follows directly from the other two. $\endgroup$ – Scooby Jul 24 '15 at 17:29
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I think we can get a better approximation of $1/\zeta^{2}\left(s\right) $. Take $\epsilon>0 $ and $\sigma\geq1+\epsilon>1 $. We recall that if $\sigma>1 $ we have $$\frac{1}{\zeta\left(s\right)}=\sum_{n\geq1}\frac{\mu\left(s\right)}{n^{s}} \tag{1} $$ where $\mu\left(s\right) $ is the Mobius function. Now if we consider the partial sum of $(1) $ we have, by partial summation, $$ \sum_{n\leq N}\frac{\mu\left(n\right)}{n^{s}}=M\left(N\right)N^{-s}+s\int_{1}^{N}M\left(u\right)u^{-s-1}du $$ where $M\left(N\right)=\sum_{n\leq N}\mu\left(n\right) $ is the Mertens function. By PNT, we know that $M\left(N\right)=o\left(N\right) $ as $N\rightarrow\infty $, then $$\left|\frac{1}{\zeta\left(s\right)}\right|=O\left(\left|s\right|\int_{1}^{\infty}u^{-\sigma}du\right)=O\left(\frac{\left|s\right|}{1-\sigma}\right) $$ then we have that exists some $C>0 $ such that $$\left|\frac{1}{\zeta\left(s\right)}\right|\leq C\left|t\right| $$ and so $$\left|\frac{\zeta''\left(s\right)\zeta\left(s\right)-2\zeta'\left(s\right)}{\zeta^{2}\left(s\right)}\right|\leq C_{1}\left|t\right|^{3}. $$

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  • $\begingroup$ cool but this is useful only for $s = 1+it$ because for $s > 1$ : $\frac{1}{\zeta(s)} = \sum \mu_n n^{-s}$, $ |1/\zeta(s)| < \sum n^{-\sigma} = \zeta(\sigma) = \mathcal{O}(1/(\sigma-1))$ $\endgroup$ – reuns Jul 18 '15 at 8:35
  • $\begingroup$ $\mathcal{O}( |t|^0)$ ? it's better than $\mathcal{O}(|t|)$, but again yours is useful for $\Re(s) = 1$ (and $\Re(s) > \sigma_0$) $\endgroup$ – reuns Jul 18 '15 at 8:50
  • $\begingroup$ Yes, you're right :) $\endgroup$ – Marco Cantarini Jul 18 '15 at 8:59
  • $\begingroup$ and I'm wondering if just with the PNT we can say that $1/|\zeta(1+it)| = \mathcal{O}(|t|^\epsilon)$ $\endgroup$ – reuns Jul 18 '15 at 9:01

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