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I guess this is a known result but I could not find it on the Internet. Consider these equations formed from the reciprocals of the divisors of $n$ raised to a complex number $s=a+ib$ :

$\displaystyle 1$

$\displaystyle 1 + \frac{1}{2^s}=0$

$\displaystyle 1 + \frac{1}{3^s}=0$

$\displaystyle 1 + \frac{1}{2^s} + \frac{1}{4^s}=0$

$\displaystyle 1 + \frac{1}{5^s}=0$

$\displaystyle 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{6^s}=0$

$\displaystyle 1 + \frac{1}{7^s}=0$

$\displaystyle 1 + \frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{8^s}=0$

$\displaystyle 1 + \frac{1}{3^s} + \frac{1}{9^s}=0$

$\displaystyle 1 + \frac{1}{2^s} + \frac{1}{5^s} + \frac{1}{10^s}=0$

$\displaystyle 1 + \frac{1}{11^s}=0$

$\displaystyle 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \frac{1}{6^s} + \frac{1}{12^s}=0$


Is it true that the zeros have real part equal to $0$ ?

Wolfram Alpha example, sums of reciprocals of divisors of 30

This Mathematica program finds some of the zeros up to the number $42$.

Clear[nn, s];
 nn = i;
 Print[s /.Table[FindRoot[Total[1/Divisors[nn]^(s)] == 0,
  {s, 5*n*I}], {n, 1, 30}]], {i, 2, 42}]

There are cases when the FindRoot command in Mathematica gives solutions that have real part not equal to zero, but these appear to be accompanied with warning messages and I have checked numerically that those solutions are not zeros.

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I don't think so; they look like they just have zeros with very nearly real part zero. The first one, for instance, also has zeros with real parts about 94.5 and 349.8. A variant of this question that I've personally wondered is if $$\sum_{n=1}^N n^{-s}$$ always has zeros with real part in $(0,1)$ (which numerically appears correct). – anon Oct 3 '11 at 9:59
It could be you are right. But by considering a different version where the reciprocals are raised to $(s-1)$ I get real part equal to $1$. That is I entered in Wolfram Alpha: Table[FindRoot[1+1/2^(s-1)+1/3^(s-1)+1/5^(s-1)+1/6^(s-1)+1/10^(s-1)+1/15^(s-1)+1‌​/30^(s-1)==0,{s,5*n*I}], {n, 1, 30}] – Mats Granvik Oct 3 '11 at 10:06
Interesting.. but the period in "1." makes me think that's still approximate. I should also mention you're specifically talking about $\sigma_{-s}(n)$ (divisor function) as a function of complex $s$. – anon Oct 3 '11 at 10:34
As a Mathematica note: DivisorSigma[-s, nn] is a simpler expression for the functions you're interested in... (edit: and I only saw @anon's comment just now.) – J. M. Oct 3 '11 at 10:36
Since you're in Mathematica: it might be useful for you to peer at contours: Partition[Table[ContourPlot[{Re[DivisorSigma[-(x + I y), k]] == 0, Im[DivisorSigma[-(x + I y), k]] == 0}, {x, -10, 10}, {y, -10, 10}], {k, 2, 41}], 5] // Grid – J. M. Oct 3 '11 at 10:45
up vote 2 down vote accepted

If $s=0$, the property fails hence one can assume without loss of generality that $s\ne0$ and that the integer $n\geqslant2$ is the product of some $p_j^{k_j}$. Define $Z=2i\pi\mathbb Z$ as the set of integer multiples of $2i\pi$.

First assume that $p_j^s\ne1$ for every $j$. Then, the sum you are interested in is $$ S_n(s)=\prod_j\left(\sum_{t=0}^{k_j}p_j^{-st}\right)=\prod_j\frac{1-p_j^{-s(k_j+1)}}{1-p_j^{-s}}. $$ The zeros $s$ of $S_n$ solve one of the equations $$ p_j^{s(k_j+1)}=1. $$ This happens if and only if $s(k_j+1)\log p_j$ is in $Z$.

On the other hand, $p_j^s=1$ for some $j$ means that $s\log p_j$ is in $Z$. For every such $j$, one should replace the contribution of $p_j$ in $S_n(s)$ by $k_j+1\ne0$.

One is left with the condition that there exists at least one $j$ such that $s\log p_j$ is not in $Z$ and $s(k_j+1)\log p_j$ is in $Z$. That is, there exists $j$ and an integer $\ell$ which is not zero modulo $k_j+1$, such that $$ s=\frac{2i\pi\ell}{(k_j+1)\log p_j}. $$ Note that since the $\log p_j$ are linearly independent over the rationals, this can happen for at most one prime number $p_j$ dividing $n$.

In particular the real part of $s$ must be zero.

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Oh duh, just factor $\sigma_{-s}(n)$. I should have thought of that. – anon Oct 3 '11 at 13:54
@anon, provided you are the anon who found zeros with real parts about 94.5 and 349.8, I am curious to know if you made a mistake, or if I misunderstood what you said, or what. – Did Oct 3 '11 at 13:58
I changed OP's example to just $1+1/2^s$. Either I misunderstood the code or Alpha screwed up big time. – anon Oct 3 '11 at 14:01
@anon, did you write 1/2(s) instead of 1/2^s? – Did Oct 3 '11 at 14:09
M.SE keeps mutilating the link, so just type in Table[FindRoot[1+1/2^(s)==0,{s,5*n*I}],{n,1,30}] and you should see this (.png screengrab). – anon Oct 4 '11 at 0:14

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