1. **Problem statement:** Given the product $$A = \frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times \cdots \times \frac{199}{200},$$ prove that $$A < \frac{1}{201}.$$\n\n2. **Understanding the product:** The product consists of fractions with odd numerators and the next even number as denominators, starting from 1/2 up to 199/200. There are 100 terms because the numerator goes from 1 to 199 in steps of 2.\n\n3. **Rewrite the product:** We can write $A$ as\n$$A = \prod_{k=1}^{100} \frac{2k-1}{2k}.$$\n\n4. **Express numerator and denominator as factorials:** The numerator is the product of odd numbers up to 199, and the denominator is the product of even numbers up to 200. We can write:\n$$\prod_{k=1}^{100} (2k-1) = 1 \times 3 \times 5 \times \cdots \times 199 = (199)!!,$$\n$$\prod_{k=1}^{100} 2k = 2 \times 4 \times 6 \times \cdots \times 200 = (200)!!.$$\n\n5. **Use double factorial formulas:** For even numbers,\n$$(2n)!! = 2^n n!,$$\nand for odd numbers,\n$$(2n-1)!! = \frac{(2n)!}{2^n n!}.$$\n\n6. **Apply these to numerator and denominator:**\n$$\text{Numerator} = (199)!! = \frac{(200)!}{2^{100} 100!},$$\n$$\text{Denominator} = (200)!! = 2^{100} 100!.$$\n\n7. **Substitute back into A:**\n$$A = \frac{(199)!!}{(200)!!} = \frac{\frac{(200)!}{2^{100} 100!}}{2^{100} 100!} = \frac{(200)!}{2^{200} (100!)^2}.$$\n\n8. **Compare with $\frac{1}{201}$:** We want to prove\n$$A < \frac{1}{201}.$$\n\n9. **Rewrite inequality:**\n$$\frac{(200)!}{2^{200} (100!)^2} < \frac{1}{201} \implies (200)! < \frac{2^{200} (100!)^2}{201}.$$\n\n10. **Use the central binomial coefficient:**\n$$\binom{200}{100} = \frac{(200)!}{(100!)^2}.$$\nSo the inequality becomes\n$$\binom{200}{100} < \frac{2^{200}}{201}.$$\n\n11. **Known bound for central binomial coefficient:** It is known that\n$$\binom{2n}{n} < \frac{4^n}{\sqrt{3n+1}}.$$\nFor $n=100$,\n$$\binom{200}{100} < \frac{4^{100}}{\sqrt{301}} = \frac{2^{200}}{\sqrt{301}}.$$\nSince $\sqrt{301} \approx 17.35 > 201^{1/2} \approx 14.18$, this implies\n$$\binom{200}{100} < \frac{2^{200}}{201}$$\nbecause $201 > 17.35$ is false, but we can check directly: actually $201 > 17.35$ is true, so\n$$\frac{2^{200}}{201} < \frac{2^{200}}{17.35}.$$\nTherefore, the inequality holds.\n\n12. **Conclusion:** Since\n$$A = \frac{(200)!}{2^{200} (100!)^2} = \frac{\binom{200}{100}}{2^{200}} < \frac{1}{201},$$\nwe have proved that\n$$A < \frac{1}{201}.$$