A curious prime fraction formula

Mathematica, 32 bytes

1##&@@(1-2/(Prime@Range@#^2+1))&

An unnamed function that takes integer input and returns the actual fraction.

This uses the fact that (p2-1)/(p2+1) = 1-2/(p2+1). The code is then golfed thanks to the fact that Mathematica threads all basic arithmetic over lists. So we first create a list {1, 2, ..., n}, then retrieve all those primes and plug that list into the above expression. This gives us a list of all the factors. Finally, we multiply everything together by applying Times to the list, which can be golfed to 1##&.

Alternatively, we can use Array for the same byte count:

1##&@@(1-2/(Prime~Array~#^2+1))&

M, 9 bytes

RÆN²‘İḤCP

Try it online!

Trivia

Meet M!

M is a fork of Jelly, aimed at mathematical challenges. The core difference between Jelly and M is that M uses infinite precision for all internal calculations, representing results symbolically. Once M is more mature, Jelly will gradually become more multi-purpose and less math-oriented.

M is very much work in progress (full of bugs, and not really that different from Jelly right now), but it works like a charm for this challenge and I just couldn't resist.

How it works

RÆN²‘İḤCP  Main link. Argument: n

R          Range; yield [1, ..., n].
 ÆN        Compute the kth primes for each k in that range.
   ²‘      Square and increment each prime p.
     İ     Invert; turn p² + 1 into the fraction 1 / (p² + 1).
      Ḥ    Double; yield 2 / (p² + 1).
       C   Complement; yield 1 - 2 / (p² + 1).
        P  Product; multiply all generated differences.

Python 2, 106 bytes

from fractions import*
n=input()
F=k=P=1
while n:b=P%k>0;n-=b;F*=1-Fraction(2*b,k*k+1);P*=k*k;k+=1
print F

The first and fourth lines hurt so much... it just turned out that using Fraction was better than multiplying separately and using gcd, even in Python 3.5+ where gcd resides in math.

Prime generation adapted from @xnor's answer here, which uses Wilson's theorem.