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Question Number 193785 by aba last updated on 19/Jun/23

$$provethat$$$$\frac{1}{2}\times \frac{3}{4}\times \frac{5}{6}\times \frac{7}{8}\times \frac{9}{10}\times ...\times \frac{99}{100}<\frac{1}{10}$$

Answered by MM42 last updated on 20/Jun/23

$$a=\frac{1}{2}\times \frac{3}{4}\times \frac{5}{6}...\times \frac{99}{100}<$$$$<\frac{2}{3}\times \frac{4}{5}\times \frac{6}{7}\times ...\times \frac{98}{99}\times \frac{100}{101}=$$$$=\frac{1}{\frac{3}{2}\times \frac{5}{4}\times \frac{7}{6}\times ...\times \frac{99}{98}\times \frac{101}{100}}$$$$\Rightarrow a<\frac{1}{101a}\Rightarrow a<\frac{1}{\sqrt{101}}<\frac{1}{10}✓$$

Commented by aba last updated on 20/Jun/23

$$\mathrm{perfect}$$

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