\(\frac{1.3.5...79}{2.4.6...80}\)= \(\frac{1.3.5...79}{\left(1.2\right).\left(2.2\right).\left(3.2\right)...\left(40.2\right)}\).\(\frac{1.3.5...79}{\left(1.2.3.4...40\right).\left(2.2.2.2...2.2\right)}\)=\(\frac{1.3.5...79}{\left(1.3.5...39\right).\left(2.4.6...40\right).2^{40}}\)<1/9