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Part of 1968 Polish MO Finals

Problems(1)

sum n /3 *5*7*...*(2n+1) <1/2 - 1968 Polish Mo finals p2

Source:

8/27/2024
Prove that for every natural nn 13+23⋅5+33⋅5⋅7+...+n3⋅5⋅7⋅...⋅(2n+1)<12.\frac{1}{3} + \frac{2}{3\cdot 5} + \frac{3}{3 \cdot 5 \cdot 7} + ...+ \frac{n}{3 \cdot 5 \cdot 7 \cdot ...\cdot (2n+1)} < \frac{1}{2}.
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