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Numerical evidence suggests that:

$$\displaystyle F(s):= \lim_{N \to \infty}\, \ln^s\left(p_N\right)\, \prod_{n=1}^N \left(\dfrac{\left(p_n-1\right)^s}{p_n^s-1} -\frac{1}{p_n^s}\right)$$

with $p_n$ is the n-th prime number, converges for all $\Re(s) > \frac12$.

Note that for $s=1$ the function reduces to the well known limit for $e^{-\gamma}$ (see here).

Could this be proven?

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  • $\begingroup$ How do you compute such complicated products over primes? $\endgroup$ – joro Nov 29 '15 at 11:59
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    $\begingroup$ Joro, I used Pari/GP on SageMathCloud. Here is the code: gp("tesp(s,v)=(log(precprime(v)))^s*prodeuler(p=1,v,((p-1)^s)/(p^s-1)-1/(p^(s)))") Easily calculates up to 10^7 primes. $\endgroup$ – Agno Nov 29 '15 at 12:58
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    $\begingroup$ In your code, have you tried s=1/2+10.0^(-8) and v=10^n as $n$ grows? I get steady increase, for $n=8$ get $38.3...$, while for n=6 it is 28.72...? If it diverges very slowly like log(log(N)) you wouldn't detect it this way AFAICT. Btw, you may have precision issues, not sure. $\endgroup$ – joro Nov 29 '15 at 15:50
  • $\begingroup$ Joro, you are right that very near to $s=\frac12$, it becomes quite difficult to get reliable numerical indications for convergence (that is slow anyway). What surprised me is that although $s=\frac12$ clearly diverges, the imaginary values $\frac12 \pm t i$ do appear to converge. Do you observe the same? $\endgroup$ – Agno Nov 29 '15 at 16:59
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As you mention, $F(1)=e^{-\gamma}$, so $$ \begin{align*} e^{s\gamma} F(s)=\frac{F(s)}{F(1)^s}&=\frac{\lim_{x\to\infty} \log^s x \prod_{p\leq x} \frac{p^s(p-1)^s-p^s+1}{p^s(p^s-1)}}{\lim_{x\to\infty} \log^s x \prod_{p\leq x} \frac{(p-1)^s}{p^s}}\\ &= \prod_{p}\frac{p^s(p-1)^s-p^s+1}{(p^s-1)(p-1)^s}\\ &=\prod_{p}\left(1-\frac{p^s-(p-1)^s-1}{(p^s-1)(p-1)^s}\right). \end{align*} $$ This product converges because $$ \frac{p^s-(p-1)^s-1}{(p^s-1)(p-1)^s}=\frac{s}{p^{s+1}}+O\left(\frac{1}{p^{2s}}\right)+O\left(\frac{1}{p^{s+2}}\right) $$ as $p\to\infty$.

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If you write the general terms in terms of $x = 1/p_n,$ when $x\>1/2,$ the expansion in $x$ starts as $1-s x + O(x^{2 s})\dots,$ so your limit converges for $s>1/2.$ For $s=1/2,$ the expansion starts with $1+x/2,$ so your limit blows up (the product of the first $n$ terms, including the log fudge factor, should be of order of $\log n.$

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  • $\begingroup$ Many thanks both Igor and Julian. Both your answers are much appreciated! $\endgroup$ – Agno Nov 29 '15 at 21:22

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