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Question Number 193848 by Mingma last updated on 21/Jun/23

Answered by MM42 last updated on 21/Jun/23

$$\frac{1}{2}\times \frac{3}{4}\times \frac{5}{6}\times ...\times ?\times \frac{2024}{2025}$$

Answered by witcher3 last updated on 21/Jun/23

$$\mathrm{let}\underset{\mathrm{k}=1}{\prod ^{\mathrm{m}}}\frac{(2\mathrm{k}-1)}{2\mathrm{k}}=\underset{\mathrm{k}=1}{\prod ^{\mathrm{m}}}(1-\frac{1}{2\mathrm{k}})$$$$1-\mathrm{x},$$$${\mathrm{e}}^{-\mathrm{x}}=1-\mathrm{x}+\frac{\mathrm{C}}{2}{\mathrm{x}}^2,\mathrm{c}\in {\mathbb{R}}_{+}$$$$\Rightarrow 1-\mathrm{x}⩽{\mathrm{e}}^{-\mathrm{x}}$$$$\Rightarrow \underset{\mathrm{k}=1}{\prod ^{\mathrm{m}}}(1-\frac{1}{2\mathrm{k}})⩽\underset{\mathrm{k}=1}{\prod ^{\mathrm{m}}}{\mathrm{e}}^{-\frac{1}{2\mathrm{k}}}={\mathrm{e}}^{-\frac{1}{2}\underset{\mathrm{k}=1}{\sum ^{\mathrm{m}}}\frac{1}{\mathrm{k}}}$$$$={\mathrm{e}}^{-\frac{1}{2}{\mathrm{H}}_{\mathrm{m}}}$$$${\mathrm{H}}_{\mathrm{m}}..\mathrm{harmonic}\mathrm{Numer}$$$${\mathrm{H}}_{\mathrm{m}}⩾\ln (\mathrm{m}+1)$$$$\Rightarrow -{\mathrm{H}}_{\mathrm{m}}⩽\ln \left(\frac{1}{\mathrm{m}+1}\right)$$$$\Rightarrow {\mathrm{e}}^{-\frac{1}{2}{\mathrm{H}}_{\mathrm{m}}}⩽{\mathrm{e}}^{\ln \left(\frac{1}{\sqrt{\mathrm{m}+1}}\right)}$$$$\Rightarrow \underset{\mathrm{k}=1}{\prod ^{\mathrm{m}}}(1-\frac{1}{2\mathrm{k}})⩽\frac{1}{\sqrt{\mathrm{m}+1}}$$$$\mathrm{m}=1022$$$$\underset{\mathrm{k}=1}{\prod ^{\mathrm{m}}}(1-\frac{1}{2\mathrm{k}})=\frac{1.3......2023}{2........2024}⩽\frac{1}{\sqrt{1022+1}}$$$$⩽\frac{1}{\sqrt{1023}}⩽\frac{1}{32}$$

Commented by MM42 last updated on 21/Jun/23

$$a=\frac{1}{2}\times \frac{3}{4}\times ...\times \frac{2n-1}{2n}$$$$<\frac{2}{3}\times \frac{4}{5}\times ...\times \frac{2n}{2n-1}\times \frac{2n}{2n+1}$$$$\frac{1}{\frac{3}{2}\times \frac{5}{4}\times ...\times \frac{2n+1}{2n}}=\frac{1}{a\times (2n+1)}$$$$\Rightarrow a<\frac{1}{\sqrt{2n+1}}$$$$therearetobeamistakinthe$$$$textofthequestion.$$

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