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Question-193848




Question Number 193848 by Mingma last updated on 21/Jun/23
Answered by MM42 last updated on 21/Jun/23
(1/2)×(3/4)×(5/6)×...× ? ×((2024)/(2025))
$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{5}}{\mathrm{6}}×…×\:?\:×\frac{\mathrm{2024}}{\mathrm{2025}}\: \\ $$
Answered by witcher3 last updated on 21/Jun/23
let Π_(k=1) ^m (((2k−1))/(2k))=Π_(k=1) ^m (1−(1/(2k)))  1−x,  e^(−x) =1−x+(C/2)x^2 ,c∈R_+   ⇒1−x≤e^(−x)   ⇒Π_(k=1) ^m (1−(1/(2k)))≤Π_(k=1) ^m e^(−(1/(2k))) =e^(−(1/2)Σ_(k=1) ^m (1/k))   =e^(−(1/2)H_m )   H_m ..harmonic Numer    H_m ≥ln(m+1)  ⇒−H_m ≤ln((1/(m+1)))  ⇒e^(−(1/2)H_m ) ≤e^(ln((1/( (√(m+1))))))   ⇒Π_(k=1) ^m (1−(1/(2k)))≤(1/( (√(m+1))))  m=1022  Π_(k=1) ^m (1−(1/(2k)))=((1.3......2023)/(2........2024))≤(1/( (√(1022+1))))  ≤(1/( (√(1023))))≤(1/(32))
$$\mathrm{let}\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{m}} {\prod}}\frac{\left(\mathrm{2k}−\mathrm{1}\right)}{\mathrm{2k}}=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{m}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2k}}\right) \\ $$$$\mathrm{1}−\mathrm{x}, \\ $$$$\mathrm{e}^{−\mathrm{x}} =\mathrm{1}−\mathrm{x}+\frac{\mathrm{C}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} ,\mathrm{c}\in\mathbb{R}_{+} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{x}\leqslant\mathrm{e}^{−\mathrm{x}} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{m}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2k}}\right)\leqslant\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{m}} {\prod}}\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2k}}} =\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{m}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}} \\ $$$$=\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{H}_{\mathrm{m}} } \\ $$$$\mathrm{H}_{\mathrm{m}} ..\mathrm{harmonic}\:\mathrm{Numer} \\ $$$$\:\:\mathrm{H}_{\mathrm{m}} \geqslant\mathrm{ln}\left(\mathrm{m}+\mathrm{1}\right) \\ $$$$\Rightarrow−\mathrm{H}_{\mathrm{m}} \leqslant\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{m}+\mathrm{1}}\right) \\ $$$$\Rightarrow\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{H}_{\mathrm{m}} } \leqslant\mathrm{e}^{\mathrm{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{m}+\mathrm{1}}}\right)} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{m}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2k}}\right)\leqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{m}+\mathrm{1}}} \\ $$$$\mathrm{m}=\mathrm{1022} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{m}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2k}}\right)=\frac{\mathrm{1}.\mathrm{3}……\mathrm{2023}}{\mathrm{2}……..\mathrm{2024}}\leqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{1022}+\mathrm{1}}} \\ $$$$\leqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{1023}}}\leqslant\frac{\mathrm{1}}{\mathrm{32}} \\ $$
Commented by MM42 last updated on 21/Jun/23
a=(1/2)×(3/4)×...×((2n−1)/(2n))  <(2/3)×(4/5)×...×((2n)/(2n−1))×((2n)/(2n+1))  (1/((3/2)×(5/4)×...×((2n+1)/(2n))))=(1/(a×(2n+1)))  ⇒a<(1/( (√(2n+1))))   there are to be a mistak  in the   text of the question.
$${a}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×…×\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}} \\ $$$$<\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{4}}{\mathrm{5}}×…×\frac{\mathrm{2}{n}}{\mathrm{2}{n}−\mathrm{1}}×\frac{\mathrm{2}{n}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{5}}{\mathrm{4}}×…×\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}}}=\frac{\mathrm{1}}{{a}×\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$\Rightarrow{a}<\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{n}+\mathrm{1}}}\: \\ $$$${there}\:{are}\:{to}\:{be}\:{a}\:{mistak}\:\:{in}\:{the}\: \\ $$$${text}\:{of}\:{the}\:{question}. \\ $$

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