I am currently studying the following inequality involving the square of the modulus of a specific Dirichlet polynomial:

$$\left( \sum_{1}^{N}\frac{1}{n} \right)^2 \ \ - \ \left| \sum_{1}^{N}\frac{(-1)^{n-1}}{n^{1/2+it}} \right| ^2 \ > \ \ 0$$ which, for arbitrary combinations of $N$ and $t$, is in general false. Just take the example $t=749.2$ plotting the above difference for $1<N<10000$ will show that it is negative up to about $N=3100$ (the exact value might be affected by the numerical accuracy of the particular Math SW tool). The inequality appears instead to hold true for $N$ greater than that. I then wondered whether there might exist simple functions $N(t)$ such that the above inequality is always satisfied. I played quite a lot with numerical simulations by first trying the very simple $N(t) = \lceil t\rceil^2 $ $\Rightarrow $ I was unable to find any violation. But of course, this is just an "experimental" approach. I wonder whether anybody may suggest specific references useful for the theoretical study, verification, or rebuttal, of the above inequality for particular $N(t)$ functions.

Many thanks.


1 Answer 1


I am assuming $t$ is real. Since $$ \left|\frac{(-1)^{n-1}}{n^{1/2+it}}+\frac{(-1)^{n+1-1}}{(n+1)^{1/2+it}} \right| \le|1/2+it|\max_{0\le s\le1}\left|\frac1{(n+s)^{3/2+it}}\right| =\sqrt{t^2+1/4}\,\frac1{n^{3/2}}, $$ the series $\sum_{1}^\infty\frac{(-1)^{n-1}}{n^{1/2+it}}$ converges, whereas the harmonic series diverges. So indeed, for each real $t$ there is some natural $N(t)$ such that the inequality in question holds for all natural $N\ge N(t)$.

Addition in response to the modification of the question: More specifically, $$\left|\sum_{1}^N\frac{(-1)^{n-1}}{n^{1/2+it}}\right| \le1+\sqrt{t^2+1/4}\,\sum_{j=1}^\infty\frac1{(2j-1)^{3/2}} =1+c\sqrt{t^2+1/4},$$ where $c:=(1-\frac1{2\sqrt2})\zeta(3/2)<1.7$, whereas $H_N>\gamma+\ln N$. So, the inequality in question holds for all natural $N\ge N(t)$, where $N(t):=\exp\{1-\gamma+c\sqrt{t^2+1/4}\}$ and $\gamma$ is the Euler constant.

Another addition: Better bounds Luca asked if the above bound on $N(t)$ can be improved, to get something closer to $\lceil t\rceil^2 $.

1. One such improvement is an easy modification of the above reasoning. Indeed, let $t\to\infty$ and let $N>A:=\lceil t\rceil$. Then \begin{equation*} \left|\sum_{1}^N\frac{(-1)^{n-1}}{n^{1/2+it}}\right| \le\left|\sum_{1}^A\frac{1}{n^{1/2}}\right| +\left|\sum_{A+1}^N\frac{(-1)^{n-1}}{n^{1/2+it}}\right| \ll A^{1/2}+\frac{\sqrt{t^2+1/4}}{A^{1/2}}\sim 2t^{1/2}. \end{equation*} So, the inequality in question holds for all natural $N\ge N(t)$, where $N(t):=\exp\{c_1+c_2 t^{1/2}\}$, where $c_1$ and $c_2$ are some positive real constants.

Further improvements seem very difficult:

2. Again, let $t\to+\infty$. The Lindelöf hypothesis -- which is implied by the Riemann hypothesis (\url{https://en.wikipedia.org/wiki/Lindel%C3%B6f_hypothesis}, \url{https://terrytao.wordpress.com/tag/riemann-zeta-function/}) -- states that $|\zeta(1/2+it)|\ll t^\epsilon$ for any $\epsilon>0$. Since $\eta(z):=\sum_{1}^\infty\frac{(-1)^{n-1}}{n^z}=\zeta(z)(1-2^{1-z})$, the Lindelöf hypothesis implies $|\eta(1/2+it)|\ll t^\epsilon$ for any $\epsilon>0$. So, assuming $N>t^{2-2\epsilon}$, we have \begin{equation*} \left|\sum_{1}^N\frac{(-1)^{n-1}}{n^{1/2+it}}\right| \le\left|\sum_{1}^\infty\frac{(-1)^{n-1}}{n^{1/2+it}}\right| +\left|\sum_{N+1}^\infty\frac{(-1)^{n-1}}{n^{1/2+it}}\right| \ll t^\epsilon+\frac{\sqrt{t^2+1/4}}{N^{1/2}}\ll t^\epsilon. \end{equation*} So, modulo the Lindelöf hypothesis, the inequality in question holds for all natural $N\ge N(t)$, where $N(t):=\exp\{c_1+c_2 t^\epsilon\}$, where $c_1$ and $c_2$ are some positive real constants.

3. By plotting with Mathematica, I have not been able to find a counterexample to the inequality \begin{equation*} |\eta(1/2+it)|\le2\ln t \tag{*} \end{equation*} for any $t\in[2,10^7]$; the closest to $2$ value of $|\eta(1/2+it)|/\ln t$ for such $t$ seems to occur near $t=6.5\times10^5$. \ One may conjecture that $(*)$ holds for all real $t\ge2$.

Again, let $t\to+\infty$. Let $c_*=0.86568\dots$ be the positive root of the equation $c_*^2 e^{c_*-\gamma}=1$, and let $N(t):=bt^2$, where $b$ is any fixed real number greater than $e^{c_*-\gamma}=1/c_*^2=1.3343\dots$, so that $b^{-1/2}<c_*$. Assume that $(*)$ holds for all large enough real $t$. Then for all large enough $t>0$ and all $N\ge N(t)$ we have \begin{equation*} \left|\sum_{1}^N\frac{(-1)^{n-1}}{n^{1/2+it}}\right| \le\left|\sum_{1}^\infty\frac{(-1)^{n-1}}{n^{1/2+it}}\right| +\left|\sum_{N+1}^\infty\frac{(-1)^{n-1}}{n^{1/2+it}}\right| \le2\ln t+\frac{\sqrt{t^2+1/4}}{N^{1/2}}\,(1+o(1)) \end{equation*} \begin{equation*} \le2\ln t+\frac{1+o(1)}{b^{1/2}} \le2\ln t+c_* < \ln N+\gamma<H_N. \end{equation*} So, modulo conjecture $(*)$, the inequality in question holds for all natural $N\ge N(t):=1.3344t^2$.

This is still a bit short of $N(t)=\lceil t\rceil^2$. However, this may to an extent explain that "empirical" bound, $N(t)=\lceil t\rceil^2$.

However, conjecture $(*)$ is not true: according to \url{https://arxiv.org/abs/1704.06158v2}, $\limsup_{t\to\infty}|\eta(1/2+it)|/e^{c\ell(t)}>1$ for any $c\in(0,1)$, where $\ell(t):=\sqrt{\ln t\;\ln\ln\ln t\,/\,\ln\ln t}$.

It follows that necessarily $N(t)>\exp\{e^{(1-o(1))\ell(t)}\}$ for some however large $t$, and of course $\exp\{e^{(1-o(1))\ell(t)}\}$ is much greater than $t^2$ for large $t>0$. Indeed, take any $c\in(0,1)$. Then, for some however large $t$ and any natural $N\in(t^2,\exp\{e^{c\ell(t)}-3\})$,
\begin{equation*} H_N\le1+\ln N<e^{c\ell(t)}-2<|\eta(1/2+it)|-2<|\eta(1/2+it)|-2\frac t{N^{1/2}} \end{equation*} \begin{equation*} \le\left|\sum_{1}^\infty\frac{(-1)^{n-1}}{n^{1/2+it}}\right| -\left|\sum_{N+1}^\infty\frac{(-1)^{n-1}}{n^{1/2+it}}\right| \le \left|\sum_{1}^N\frac{(-1)^{n-1}}{n^{1/2+it}}\right|, \end{equation*} so that the inequality in question fails to hold.

  • $\begingroup$ Thank you Iosif for your clarification. I have now edited my question from "might exist functions $N(t)$ $\endgroup$
    – Luca
    Sep 9, 2017 at 8:34
  • $\begingroup$ Thank you Iosif for your clarification. I have now edited the typo in my question from "might exist functions $N(t)$" to "might exist $simple$ functions $N(t)$", which is what I instead meant. Indeed the series (also known as alternating zeta function series), whose $Nth$ partial sum is that Dirichlet polynomial, is well known to converge to the Dirichlet $\eta$ function. The reason why I have squared also the $Nth$ harmonic number was the hope to then be able to more easily group subsets of the total number of terms in meaningful ways, but got nowhere ... $\endgroup$
    – Luca
    Sep 9, 2017 at 8:43
  • $\begingroup$ I have now given a simple expression for $N(t)$. $\endgroup$ Sep 10, 2017 at 1:46
  • $\begingroup$ Thanks for the useful answer. Of course, if others might have additional suggestions, on how to further lower Iosif bound closer to the $N(t) = \lceil t\rceil^2 $ $\Rightarrow $ suggested by my (admittedly limited) numerical simulations, those would also be welcomed. $\endgroup$
    – Luca
    Sep 12, 2017 at 8:24
  • $\begingroup$ I have added considerations of much improved bounds. $\endgroup$ Sep 15, 2017 at 15:47

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