I would like to know if it in the literature an approximation for $$\sum_{\rho}\frac{1}{|\rho|^2}\tag{1}$$ where the sum is over all of the non-trivial zeros of the Ramanujan's zeta function (also known as Ramanujan $\tau$-Dirichlet series). I know a similar series as Lemma 3.3 from [1], and that there are similar series involving non-trivial zeros of the Riemann zeta function in the literature. If isn't in the literature I would like how get and justify an approximation of $(1)$.

Question. Please refer the literature where is approximated $$\sum_{\rho}\frac{1}{|\rho|^2}$$ where the sum runs over all the non-trivial zeros of the Ramanujan's zeta function, and I try to search and read from the literature how was it computed. In case that it is unknown, please add your approach to calculate and justify an approximation. Many thanks.

Feel free if you want to add an optional comment/remark mentioning what convergent series involving non-trivial zeros (or convergent series involving the imaginary parts of those) of the Ramanujan's zeta function can be approximated, and that are similar to those series that are in the literature involving non-trivial zeros of the Riemann zeta function.

As reference for the Ramanujan's zeta function we know from Internet the article Tau Dirichlet Series of the encyclopedia Wolfram MathWorld, and the book [2].


[1] Kevin Ford, Zero-free Regions for the Riemann Zeta Function, Number Theory for the Millenium II, A K Peters (2002).

[2] G. H. Hardy, Ramanujan: Twelve lectures on subjects suggested by his life and work, AMS Chelsea Publishing (2002).

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    $\begingroup$ I doubt such a thing exists in the literature. Have you tried generalising Kevin Ford's proof yourself? $\endgroup$ – Peter Humphries Nov 22 '19 at 17:39
  • $\begingroup$ I didn't try it @PeterHumphries It is required the knowing of the closed-form of another convergent series and the approximation of the imaginary part of the first non-trivial zero of the Riemann zeta function. $\endgroup$ – user142929 Nov 22 '19 at 17:42
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    $\begingroup$ Yes, and both of these things should be doable in this setting. Check LMFDB.org for the first nontrivial zero and generalise the proof (via the functional equation) for the closed form sum over zeroes. $\endgroup$ – Peter Humphries Nov 22 '19 at 17:43
  • $\begingroup$ Many thanks @PeterHumphries thus I know the imaginary part of first nontrivial zero for the related $L$ function from your reference. This weekend I am going to try it using the proof by Kevin Ford and your remarks. Many thanks all for the patience and generosity. $\endgroup$ – user142929 Nov 23 '19 at 8:08

Writing the nontrivial zeros $\rho$ as $\beta+i\gamma$ with $\beta,\gamma\in\mathbb{R}$, we first observe that

$\displaystyle\sum_{\rho}\frac{1}{|\rho|^2}\leq \sum_{\gamma}\frac{1}{\gamma^2}$.

We find using the first few zeros of $L(s,\Delta)$ (Ramanujan's zeta function) as computed on LMFDB that

$\displaystyle \sum_{|\gamma|\leq 32}\frac{1}{\gamma^2}=0.05999167...$

Now, consider Lemma 5.4 of https://arxiv.org/pdf/1305.5283.pdf. In our case, we have $n=1$ and $\mathfrak{q}(f)=11$ (since the associated newform is of weight 12 and level 1). We find that for any $t\in\mathbb{R}$,

$\displaystyle N(t):=\#\{\rho=\beta+i\gamma\colon |\gamma-t|\leq 1\}\leq \frac{13}{6}(\log 11 + \log(|t|+4))+8\log(|t|+4)$.

This is a GRH-conditional bound, but if you inspect the proof, it can be made unconditional by replacing $13/6$ with $5/2$ and replacing $8$ with $35/4$. Since $\tau(n)\in\mathbb{R}$, $\beta+i\gamma$ is a nontrivial zero if and only if $\beta-i\gamma$ is a nontrivial zero by the functional equation. Thus

$\displaystyle\sum_{|\gamma|>32}\frac{1}{\gamma^2}\leq 2\sum_{j=0}^{\infty}\frac{N(33+2j)}{(32+2j)^2}\leq 2\sum_{j=0}^{\infty}\frac{\frac{5}{2}(\log 11 + \log(|33+2j|+4))+\frac{35}{4}\log(|33+2j|+4)}{(32+2j)^2},$

which is $\leq 1.8298322649...$. Thus we have

$\displaystyle\sum_{\rho}\frac{1}{|\rho|^2}\leq 1.889824.$

One could be more accurate if one computed more of the zeros of low height (or one assumed GRH). Alternatively, for more general $L$-functions, one could prove a more general version of Lemma 5.4 and not compute any zeros (at the cost of having an upper bound that could be far from the truth).

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  • $\begingroup$ Many thanks I'm going to study the details of your excellent answer in next hours. In addition, I'm sure that your post will be interesting for many other people. $\endgroup$ – user142929 Mar 9 at 9:18

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