The zeros of $f(s,a) = \sum_{n=1}^{a} (\frac{n^2 + n}{2})^{-s} $

Question

I was looking at the zeros of $$ f(s,a) = \sum_{n=1}^{a} \Big(\frac{n^2 + n}{2}\Big)^{-s} $$

for integer $a>3$ in the strip $0 < \operatorname{Re}(s) < \frac{1}{2}$, and of course of the limiting case " triangular zeta function ": $$ f(s) = \sum_{n=1}^{\infty} \Big(\frac{n^2 + n}{2}\Big)^{-s} $$ in the strip $0 < \operatorname{Re}(s) < \frac{1}{2}$.

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Of course $f(s)$ has a pole at $s = \frac{1}{2}$ and I am not sure how analytic continuation works beyond there, but just like with the truncated Riemann zeta, where the zeros of them are relevant and converging to the zeros of the Riemann zeta, we get a similar situation here. Well I assume. I'm not sure if analytic continuation to the entire complex plane is possible but I assume that $Re(s) > 0$ and this seems to work fine.

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A few observations:

• For small $a$ the zeros seem to be somewhat random in the strip $0 < \operatorname{Re}(s) < 1$, while for large $a$ we seem to get some patterns.

• For some unbounded function $0 < g(a) < a$ and $s = x + g(a)i$ with $0 < x < \frac{1}{2}$ we seem to get that the zero's of $f(s,a)$ are all the strip $\frac{1}{3} < \operatorname{Re}(s) < \frac{1}{2}$.

  • It seems the line $\operatorname{Re}(s) = \frac{1}{3}$ is attracting the zeros for large $g(a)$ but for small imaginary parts it seems it is in fact repelling.
  • So it seems that the zeros are dense on a curve (looks continuous but like lightning) within $1/3 < \operatorname{Re}(s) < 1/2$ that tends towards the line $1/3$, but is not smooth and is repelled from $1/3$ for small imaginary parts.

I assume the positions of the zeros of $f(s,a)$ converge as $a$ grows or at least the curve they are on does converge to a fixed curve/path.

Questions

1. Is there a closed form for this curve/path ? (Clarifications would be nice).
2. Can contour integrals solve this ?

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A few further observations:

• It seems from observing the behavior of the first zero of $f(s,a)$ that $f(s)$ has a zero with real part $1/3$.

• It seems that $f(s)$ has no zero's in the strip $0 < Re(s) < \frac{1}{3}$.

• And it might even have a critical line at $Re(s) = \frac{1}{3}$.

Bonus question. Is any of those 3 statements true ?

Answer 1

This is not an answer just a few comments that should help:

Note that writing $$f(s) = \sum_{n=1}^{\infty} \Big(\frac{n^2 + n}{2}\Big)^{-s}=\sum_{m\ge 1}\frac{c_m}{m^s}$$ where $c_m=1$ if $m=\frac{n^2+n}{2}$ and $c_m=0$ otherwise expresses $f$ as a Dirichlet series with abscissa of absolute convergence (and convergence since $c_m \ge 0$ being $1/2$).

In particular, a classical theorem attributed to Knopp by Jentzsch (see his paper HERE, page $236$) shows that the zeroes of $f(s,a)$ cluster on the line $1/2+it$.

The proof is not difficult considering the analytic functions $g_a(s)=\log f(s,a)/\log a$ in a small neighborhood of a point $1/2+i\tau$ where the zeroes wouldn't cluster and showing that $g_a$ are uniformly bounded (by partial summation since $1/2$ is abscissa convergence) so form a normal family; but for $\Re s >1/2$ in the neighborhood, we have $g_a(s) \to 0, a \to \infty$ (because $f(s,a)\to f(s)$ there so for a dense open set in the neighborhood where $f$ is not zero we have $\log f(s,a) \to \log f(s)$ finite complex number etc) so by analytic continuation and the above $g_a(s) \to 0$ everywhere in the neighborhood (any subsequence of $g_a$ can only converge to zero and Montel applies to show that any subsequence of $g_a$ has a normally convergent sub-subsequence) and that is easily seen to give a contradiction by considering $\Re s <1/2$ there (using that $\limsup_a \log |\sum_{m \le a} c_m|/\log a=1/2$ by the abscissa definition in terms of coefficients and partial summation we cannot have $\limsup |g_a(s)|$ identically zero in a small open eet to the left of the abscissa)

Next $f(s)=2^s(\zeta(2s)-h(s))$ and $$h(s)=\sum \frac{1}{n^s}({1/n^s-1/(n+1)^s})$$ with the series easily shown to converge absolutely when $\Re s >0$ hence the claim about analytic continuation until $\Re s >0$ and most likely this can be improved by a similar process

As for zeroes when $\Re s >1/2$ I think that the method of Montgomery in which he shows that the partials of the usual Riemann Zeta $\zeta (s)=\sum n^{-s}$ have zeroes in $\Re s >1$ can be extended here to show that $f(s,a)$ have zeroes in $\Re s >1/2$ too for all large $a$ but that's just a guess