Rational sequence converging to zero at every standard absolute value ($y_n=\frac{n!}{1+(n!{)}^2}$)

The displayed nonzero rational sequence tends to zero both in the usual real absolute value and in every p-adic absolute value. In the real absolute value, $|y_n|\le 1/n!\to 0$. For a fixed prime $p$ and $n\ge p$, the denominator is congruent to one modulo $p$, so its P-adic valuation is zero. Therefore $v_p(y_n)=v_p(n!)\to \infty$, proving $|y_n{|}_p\to 0$.

This is convergence at each fixed place; it is not uniform convergence over all primes. For each particular $n$, new prime divisors of the denominator can compensate the small values at the previously fixed places, consistently with the product formula. Thus a product identity for a single rational number does not prohibit pointwise convergence at infinitely many absolute values.