I'm recently studying some topics in analytic number theory and I have encountered results involving the infinite product $$C=\prod_{p}\left(1-\frac{1}{p(p+1)}\right)$$ where $p$ denotes calculating the product over all prime numbers.
Or more intuitively,$$C=\left(1-\frac{1}{2\cdot3}\right)\cdot\left(1-\frac{1}{3\cdot4}\right)\cdot\left(1-\frac{1}{5\cdot6}\right)\cdot\left(1-\frac{1}{7\cdot8}\right)\cdot\left(1-\frac{1}{11\cdot12}\right)\cdots$$ Mathematica can show that $C\approx0.704442.$
However, I need a high precision result, roughly accurate to at least 20 digits, so that I could use the Inverse Symbolic Calculator to make an educated guess of any possible analytic expression for $C$ in forms like $\dfrac{\zeta(2)\zeta(3)}{\zeta(6)}$.
When I use NProduct
, I get
NProduct[1 - 1/(Prime[k] (Prime[k] + 1)), {k, 1, Infinity},
AccuracyGoal -> 20, PrecisionGoal -> 20]
During evaluation of In[14]:= Prime::intpp: Positive integer argument expected in Prime[15.]. >>
During evaluation of In[14]:= Prime::intpp: Positive integer argument expected in Prime[14.]. >>
During evaluation of In[14]:= Prime::intpp: Positive integer argument expected in Prime[13.]. >>
During evaluation of In[14]:= General::stop: Further output of Prime::intpp will be suppressed during this calculation. >>
0.705024
However, when I use N[Product[]]
, the calculation was so slow that I had to abort the calculation after minutes of calculating.
N[Product[1 - 1/(Prime[k] (Prime[k] + 1)), {k, 1, 100000}], 50]
0.70444223707595873775750824971600704569326374380066
N[Product[1 - 1/(Prime[k] (Prime[k] + 1)), {k, 1, 1000000}], 50]
0.70444220359808164073982578152558851839613830115938
N[Product[1 - 1/(Prime[k] (Prime[k] + 1)), {k, 1, 10000000}], 50]
$Aborted
Is there any way to calculate products like $C$ to high precision in Mathematica? Thanks.