prove-that-1-2-3-4-5-6-7-8-9-10-99-100-gt-1-13-

Question Number 193760 by York12 last updated on 19/Jun/23

prove that (1/2)×(3/4)×(5/6)×(7/8)×(9/(10))...×((99)/(100))>(1/(13))

$${prove}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{5}}{\mathrm{6}}×\frac{\mathrm{7}}{\mathrm{8}}×\frac{\mathrm{9}}{\mathrm{10}}…×\frac{\mathrm{99}}{\mathrm{100}}>\frac{\mathrm{1}}{\mathrm{13}}\:\:\: \\ $$

Answered by MM42 last updated on 19/Jun/23

can br show : Π_(i=1) ^n ((2i−1)/(2i)) > ((11)/(10(√(4n+1)))) ⇒for n=50 to have (1/2)×(3/4)×...×((99)/(100))>((11)/(10(√(201))))>(1/(13))

$${can}\:{br}\:{show}\::\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\:\frac{\mathrm{2}{i}−\mathrm{1}}{\mathrm{2}{i}}\:>\:\frac{\mathrm{11}}{\mathrm{10}\sqrt{\mathrm{4}{n}+\mathrm{1}}} \\ $$$$\Rightarrow{for}\:\:{n}=\mathrm{50}\:{to}\:{have}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×…×\frac{\mathrm{99}}{\mathrm{100}}>\frac{\mathrm{11}}{\mathrm{10}\sqrt{\mathrm{201}}}>\frac{\mathrm{1}}{\mathrm{13}} \\ $$

Commented by York12 last updated on 19/Jun/23

$${sir}\:{where}\:{can}\:{I}\:{learn}\:{that}\: \\ $$

Commented by York12 last updated on 19/Jun/23

((11)/(10(√(4n+1)))) how could you know that I feel like there is something I do not know

$$\frac{\mathrm{11}}{\mathrm{10}\sqrt{\mathrm{4}{n}+\mathrm{1}}}\:{how}\:{could}\:{you}\:{know}\:{that}\: \\ $$$${I}\:{feel}\:{like}\:{there}\:{is}\:{something}\:{I}\:{do}\:{not}\:{know} \\ $$

Commented by MM42 last updated on 19/Jun/23

s.i→i=1⇒(1/2)>((11)/(10(√5)))⇒125>121 ✓ i.h→i=k⇒Π_(i=1) ^k ((2i−1)/(2i)) >((11)/(10(√(4k+1)))) i.h→i=k+1⇒Π_(i=1) ^(k+1) ((2i−1)/(2i)) >((11)/(10(√(4k+5)))) Π_(i=1) ^(k+1) ((2i−1)/(2i)) =Π_(i=1) ^k ((2i−1)/(2i)) ×((2k+1)/(2k+2))> ((11)/(10(√(4k+1))))×((2k+1)/(2k+2)) >^★ ((11)/(10(√(4k+5)))) ★→(2k+1)(√(4k+5))>(2k+2)(√(4k+1)) (4k^2 +4k+1)(4k+5)>4(k^2 +2k+1)(4k+1) 16k^3 +36k^2 +24k+5>16k^3 +36k^2 +24k+4 5>1 ✓ it always exists

$${s}.{i}\rightarrow{i}=\mathrm{1}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}>\frac{\mathrm{11}}{\mathrm{10}\sqrt{\mathrm{5}}}\Rightarrow\mathrm{125}>\mathrm{121}\:\checkmark \\ $$$${i}.{h}\rightarrow{i}={k}\Rightarrow\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}\:\:\frac{\mathrm{2}{i}−\mathrm{1}}{\mathrm{2}{i}}\:>\frac{\mathrm{11}}{\mathrm{10}\sqrt{\mathrm{4}{k}+\mathrm{1}}}\: \\ $$$${i}.{h}\rightarrow{i}={k}+\mathrm{1}\Rightarrow\underset{{i}=\mathrm{1}} {\overset{{k}+\mathrm{1}} {\prod}}\:\:\frac{\mathrm{2}{i}−\mathrm{1}}{\mathrm{2}{i}}\:>\frac{\mathrm{11}}{\mathrm{10}\sqrt{\mathrm{4}{k}+\mathrm{5}}}\:\: \\ $$$$\:\underset{{i}=\mathrm{1}} {\overset{{k}+\mathrm{1}} {\prod}}\:\:\frac{\mathrm{2}{i}−\mathrm{1}}{\mathrm{2}{i}}\:=\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}\:\:\frac{\mathrm{2}{i}−\mathrm{1}}{\mathrm{2}{i}}\:×\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}{k}+\mathrm{2}}> \\ $$$$\frac{\mathrm{11}}{\mathrm{10}\sqrt{\mathrm{4}{k}+\mathrm{1}}}×\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}{k}+\mathrm{2}}\:\:\overset{\bigstar} {>}\:\frac{\mathrm{11}}{\mathrm{10}\sqrt{\mathrm{4}{k}+\mathrm{5}}}\: \\ $$$$\: \\ $$$$\bigstar\rightarrow\left(\mathrm{2}{k}+\mathrm{1}\right)\sqrt{\mathrm{4}{k}+\mathrm{5}}>\left(\mathrm{2}{k}+\mathrm{2}\right)\sqrt{\mathrm{4}{k}+\mathrm{1}} \\ $$$$\left(\mathrm{4}{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{1}\right)\left(\mathrm{4}{k}+\mathrm{5}\right)>\mathrm{4}\left({k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{4}{k}+\mathrm{1}\right) \\ $$$$\mathrm{16}{k}^{\mathrm{3}} +\mathrm{36}{k}^{\mathrm{2}} +\mathrm{24}{k}+\mathrm{5}>\mathrm{16}{k}^{\mathrm{3}} +\mathrm{36}{k}^{\mathrm{2}} +\mathrm{24}{k}+\mathrm{4} \\ $$$$\mathrm{5}>\mathrm{1}\:\checkmark\:{it}\:{always}\:{exists} \\ $$$$ \\ $$

Commented by York12 last updated on 20/Jun/23

$${that}\:{is}\:{proof}\:{by}\:{inductions} \\ $$$${but}\:{sir}\:{can}\:{you}\:{recommend}\:{me}\:{a}\:{books}\:{to}\:{use} \\ $$$$,\:{I}\:{am}\:{preparing}\:{for}\:{IMO}\: \\ $$

Commented by MM42 last updated on 20/Jun/23

$${Hello}\:.{dear}\:{friend}.{i}\:{am}\:{a}\:{retired} \\ $$$${teacher}\:{frome}\:{iran}.{the}\:{books}\:{available}\:{to}\: \\ $$$${me}\:{are}\:{whith}\:{persian}\:{translation}. \\ $$$${but}\:{i}\:{will}\:{send}\:{you}\:{the}\:{names}\:{of}\:{some}\:{books}. \\ $$$${i}\:{hoopr}\:{it}\:{useful}\:{for}\:{you} \\ $$$${good}\:{luck}. \\ $$$$ \\ $$

Commented by MM42 last updated on 20/Jun/23