To prove: $$\frac{1\cdot 3 \cdot 5 \cdot ... \cdot 9999}{2\cdot 4 \cdot 6 \cdot ... \cdot 10000} < \frac{1}{100} $$
I've got to the point that the left side of the equation above is equal to $$\frac{10000!}{2^{10000} \cdot (5000!)^2}$$
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3Does this answer your question? Prove the inequality $\frac{1}{1999}<\frac{1\cdot3\cdot5\cdot7\cdots1997}{2\cdot4\cdot6\cdot8\cdots1998}<\frac{1}{44}$ – Dietrich Burde Jun 26 '23 at 15:53
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Does this answer your question? Mathematical induction problem: $\frac12\cdot \frac34\cdots\frac{2n-1}{2n}<\frac1{\sqrt{2n}}$ – Dietrich Burde Jun 26 '23 at 15:58