This question loosely builds on this one.

Take the following infinite product:

$$N(s,\chi_k)=\prod_{n=2}^{\infty} \left(\frac {1}{1-\dfrac{\chi_k(n)}{n^s}} \right)$$

with $\chi_k$ a Dirichlet character with period $k$. The product appears to have quite a few closed forms e.g.:

For principal characters:

$\displaystyle N(2,\chi_{1_1})=2$

$\displaystyle N(2,\chi_{2_1})=\frac{4}{\pi} =N(2,\chi_{4_1}),\cdots$

$\displaystyle N(2,\chi_{3_1})=\frac{3\sqrt{3}}{\pi} =N(2,\chi_{9_1}),\cdots$

$\displaystyle N(2,\chi_{6_1})=\frac{2\sqrt{3}}{\pi} =N(2,\chi_{36_1}),\cdots$

For non-principal characters:

$\displaystyle N(1,\chi_{3})=\frac{3\sqrt{3}}{2\,\pi}$

$\displaystyle N(1,\chi_{4})=\frac{2\sqrt{2}}{\pi}$

$\displaystyle N(1,\chi_{6})=\frac{\pi}{3}$

Has this function been studied before? If so, I would be grateful for a reference.

Does this product over $n$ converge for non-principal characters in the domain $\Re(s)>\frac12$ (numerical evidence suggests it does), or is this just as problematic to prove as it is for the product over primes?

  • 6
    $\begingroup$ This is easy by taking logarithms and comparing with Dirichlet $L$-functions evaluated at $s$, $2s$, $3s$, etc. More generally, the product should converge for $s$ of real part $>1/m$ if $\chi^m$ is the first power of $\chi$ that's principal. $\endgroup$ Jun 21, 2015 at 19:37
  • $\begingroup$ Thanks Noam. That sounds easier than I had expected. I do have a (probably tricky) follow-up question: could the same approach be applied for establishing convergence in the domain $\Re(s)<1$ when we use the composites ($c$) instead of $n$ in the products? My logic is that since the division of the Euler products for $n$ by the Euler products for $c$ should be equal to the Euler products for primes ($p$), the division of two converging products $n$ and $c$, would therefore "determine" the domain of convergence for the prime products. $\endgroup$
    – Agno
    Jun 21, 2015 at 21:04
  • $\begingroup$ Sorry, I don't think one can make this work, because convergence over all $n$ is so much easier that convergence over composites is (as you basically note) equivalence to convergence over primes, and we "know" that convergence over primes is really hard (equivalent to the Riemann Hypothesis for that $L$-series). $\endgroup$ Jun 21, 2015 at 21:09
  • $\begingroup$ Thanks. I was already a bit afraid that would be the case. Had hoped there was just a little more "beauty" in the composites (e.g. all the series of multiples of primes) than for the primes that truly seem "to grow like weeds among the natural numbers". $\endgroup$
    – Agno
    Jun 21, 2015 at 21:21
  • 1
    $\begingroup$ Problem is there's no natural way to sum (or multiply) over composites $c$ other than go over all $n$ and remove the primes. You can go over pairs $(a,b)$ with $c=ab$, but that counts different composites with different multiplicity so it answers a different question. $\endgroup$ Jun 21, 2015 at 22:18

1 Answer 1


Don't get numerical support for $\chi_4$ and $s$ very little above $\frac12$.

Pari session for $s=0.50001$:

? S=0.50001;su=1.0;for(n=2,10^3, su *=1/(1-sin(Pi*n/2)/n^S));su
%7 = 3.3419804579133062014076196783028182393
? S=0.50001;su=1.0;for(n=2,10^4, su *=1/(1-sin(Pi*n/2)/n^S));su
%8 = 6.0077205622478092377785314966380781906
? S=0.50001;su=1.0;for(n=2,10^5, su *=1/(1-sin(Pi*n/2)/n^S));su
%9 = 10.718834041599253876097107269419901288
? S=0.50001;su=1.0;for(n=2,10^6, su *=1/(1-sin(Pi*n/2)/n^S));su
%10 = 19.078938493200700611989930109774636399
? S=0.50001;su=1.0;for(n=2,10^7, su *=1/(1-sin(Pi*n/2)/n^S));su
%11 = 33.933440461598696086507174768942971589
? S=0.50001;su=1.0;for(n=2,10^8, su *=1/(1-sin(Pi*n/2)/n^S));su
%12 = 60.337666941234336761528052099665964235
  • $\begingroup$ Thanks Joro. Checked your calculations in another CAS and get exactly the same outcomes. Note that $s=0.50001$ still seems to converge for the prime product with $\chi_4$, however also diverges for its complementary product of composites. The latter therefore seems to be driving the divergence and maybe this has something to do with the composites product approaching poles ($\rightarrow \infty$) on the line $\Re(s)=\frac12$, whereas the prime product is expected to vanish ($\rightarrow 0$) at the same spots on that line? $\endgroup$
    – Agno
    Jun 22, 2015 at 10:32
  • $\begingroup$ @Agno chi_4 is zero for even n and -1 iff $n \equiv 3 \pmod{4}$. This has very simple closed form, why not edit with the closed form for chi_4? $\endgroup$
    – joro
    Jun 22, 2015 at 10:47
  • $\begingroup$ @Agno This question maybe is within reach. For the primes, chebyshev bias is involved and it must not vanish to zero because of too much bias... $\endgroup$
    – joro
    Jun 22, 2015 at 11:10

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