Let $s=x+iy$ the complex variable. We define the following arithmetical function for $\Re s>1$ $$\sigma^s(N)=\sum_{n\mid N}n^s,$$ (thus that satisfies $\sigma^s(N)=N^s\sum_{n=1}^N 1/n^s-N^s\sum_{\substack{n\nmid N\\1<n<N}}n^s$, a similiar identity that satisfied the sum of divisors function) and the partial sum of the Riemann zeta function (this function thus convergent for $\Re s>1$ and written as $\zeta(s)=\sum_{n=1}^\infty n^{-s}$) $$H_N^s=\sum_{n=1}^N\frac{1}{n^s},$$ if I understand well are the so called generalized harmonic numbers.

On the other hand one knows Kaneko's claim and its relationship with Lagarias equivalence to the Riemann Hypothesis (see page 542 of Lagarias, An Elementary Problem Equivalent to the Riemann Hypothesis American Mathematical MONTHLY June-July 2002).

Question. Does make sense that for each fixed complex number with $\Re s>1$ $$|\sigma^s(N)|<\left|e^{H_N^s}\right|\cdot\log \left|H_N^s\right|$$ holds for all integers $N\geq N_0=N_{0}(s)$? Here we are using the complex modulus and the natural logarithm for reals numbers, and we are denoting with $N_{0}(s)$ that such integer $N_0$ is depending on $s$, which has $\Re s>1$ as was said. Thanks in advance.

Thus I am asking if you can discuss the existence of such $N_0$ is right for each $s$ with $\Re s>1$: provide us a proof of such identity or well disprove such claim.

Additionally, only if you want, thus it is optional, you can argue if such identity has meaning or can improve it to provide us a more interesting question and, as a second discussion also optinal, you can discuss if has meaning the limit $\lim{s\to 1^{+}}$ of previous identity and if has relationship with Lagarias or Kaneko. If you do a contribution about this last paragraph I am saying only a few words, but your calculations are welcome with the purpose to be the best reference in this site.

  • $\begingroup$ You can add your answer with different notation. About the notation, for the real part of a complex number as the letter sigma, currently my beliefs is that behind this notation there was intentionality. And with respect the content of the Question, currently my belief is that there is meaning in the relation between the sum of divisors function and how the Riemann Zeta function diverges to its pole. But are only beliefs, I don't know provide us a explanation. $\endgroup$ – user243301 Aug 19 '16 at 8:43
  • $\begingroup$ I hope that there are no more typos, I wrote different expressions to emphasize the right definitions. Also I've written $\sigma^s(n)$ when the right should be $\sigma_s(n)$ for a generalized divisor function ! Thanks all users. $\endgroup$ – user243301 Aug 19 '16 at 8:48
  • $\begingroup$ prove the Nicolas/Robin/Lagarias criterion before trying to generalize it $\endgroup$ – reuns Aug 19 '16 at 9:13
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    $\begingroup$ Notice that the average size of $\sigma^s(N)$ is $\frac{\zeta(s+1)}{s+1} N^s$, while $H_N^s$ is a bounded sequence. $\endgroup$ – sometempname Aug 20 '16 at 9:09
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    $\begingroup$ I'm not convinced Lagarias crtierion is a so interesting criterion for RH, but looking at its derivation is very learnful. Can you sketch a proof of its derivation ? $\endgroup$ – reuns Sep 20 '16 at 19:03

There is a sketch derivation of the Robin criterion. Once you got each of this steps, you can easily see if it can be generalized.

Let $a = \sup_\rho Re(\rho)$ such that $\zeta(s)$ has no zero $Re(\rho) > a \ge 1/2$ (the Riemann hypothesis is $a= 1/2$).

$\sigma(n) = \sum_{d | n} d$ is multiplicative : $\sigma(n) = \prod_{p^k \| n} \sigma(p^k)$. And $\sigma(p^k) = \sum_{m=0}^k p^m = \frac{p^{k+1}-1}{p-1}$, so the maximum of $\frac{\sigma(n)}{n}$ is reached when $n$ is a priomorial $N_x = \prod_{p < x} p$ : $$\frac{\sigma(N_x)}{N_x} = \prod_{p < x} 1+\frac{1}{p}$$ Taking the logarithm of this and using $$\sum_{p < x} \frac{1}{p} = M + \ln \ln x + \mathcal{O}(x^{-1+a+\epsilon})$$

and $$\ln (\prod_{p < x} 1+\frac{1}{p}) = \sum_{p < x} \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k p^k} = \sum_{p < x} \frac{1}{p} + L + \mathcal{O}(x^{-1/2+\epsilon})$$ we have $$\frac{\sigma(N_x)}{N_x} = e^{\gamma + \ln \ln x + \mathcal{O}(x^{-1+a+\epsilon})}$$ where $\gamma = M+L$, $\ \ $ $M$ being the Meissel Mertens constant and $L = \sum_p \sum_{k=2}^\infty \frac{(-1)^{k+1}}{k p^k}$

Then, use the prime number theorem $$\sum_{p < x} \ln p = x + \mathcal{O}(x^{a+\epsilon})$$ so that $\ln N_x = \sum_{p < x} \ln p= x+\mathcal{O}(x^{a+\epsilon})$ i.e. $$x = \ln N_x + \mathcal{O}((\ln N)^{a+\epsilon})$$ and you get $$\frac{\sigma(N_x)}{N_x} = e^{\gamma + \ln \ln \ln N_x + \mathcal{O}((\ln N_x )^{-1+a+\epsilon})} = e^{\gamma} \ln \ln N_x (1+\mathcal{O}((\ln N_x )^{-1+a+\epsilon}))$$ i.e. $$\frac{\sigma(n)}{n} < e^{\gamma} (1+C (\ln n )^{-1+a+\epsilon})) \ln \ln n$$ for some constant $C$ depending on $a$.

Nicolas, Robin and Lagarias showed that $C = 0$ (for $n$ large enough) is equivalent to $a = 1/2$, the Riemann hypothesis.

  • $\begingroup$ Very thanks much for your detailed answer, is incredible, I need to study a lot. $\endgroup$ – user243301 Aug 20 '16 at 10:33

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