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. 



critical line inequality concerning the square of the modulus of a Dirichlet polynomial
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. 


