Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 99975 by Dwaipayan Shikari last updated on 24/Jun/20

((1/2))^(((1/3))^((1/4)....∞) ) =?

$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}....\infty} } =? \\ $$

Answered by bachamohamed last updated on 24/Jun/20

solution: on a a_n =((1/2))^(((1/3))^(((1/4))^(........((1/(n+1)))) ) ) . a_1 =((1/2)) . a_2 =((1/2))^(((1/3))) =e^((1/3)ln(1/2)) a_3 =((1/2))^(((1/3))^(((1/4))) ) =e^((1/4)ln(((1/2))^(((1/3))) )) =e^((1/4)ln(a_2 )) =e^((1/4)ln(e^((1/3)ln((1/2))) )) =e^((2/(4!))ln((1/2))) a_4 =((1/2))^(((1/3))^(((1/4))^(((1/5))) ) ) =e^((1/5)ln(((1/2))^(((1/3))^(((1/4))) ) )) =e^((1/5)ln(a_3 )) =e^((1/5)ln(e^((2/(4!))ln((1/2))) )) =e^((2/(5!))ln((1/2))) a_5 =((1/2))^(((1/3))^(((1/4))^(((1/5))^(((1/6))) ) ) ) =e^((1/6)ln(a_4 )) =e^((1/6)ln(e^((2/(5!))ln((1/2))) )) =e^((2/(6!))ln((1/2))) .... ... a_n =((1/2))^(((1/3))^(.....((1/(n+1)))) ) =e^((1/((n+1)))ln(a_(n−1) )) =e^((1/((n+1)))ln(e^((2/(n!))ln((1/2))) )) =e^((2/((n+1)!))ln((1/2))) et :n>0 (n+1)<(n+1)!<(n+1)^((n+1)) ⇒ n>0 (2/((n+1)))>(2/((n+1)!))>(2/((n+1)^((n+1)) )) n>0 ((1/2))^(((2/(n+1)))) <((1/2))^(((2/((n+1)!)))) < ((1/2))^(2/((n+1)^((n+1)) )) ⇒ n>0 . ((1/2))^(((2/(n+1)))) <a_n =e^((2/((n+1)!))ln((1/2))) <((1/2))^(2/((n+1)^((n+1)) )) donc : lim_(n→∞) ((1/2))^(2/((n+1))) < lim_(n→∞) a_n <lim_(n→∞) ((1/2))^(2/((n+1)^((n+1)) )) ⇒ n>0 1<lim_(n→∞) a_n <1 en fin:( lim_(n→∞) a_n =1)

$$\boldsymbol{{solution}}:\: \\ $$$$\boldsymbol{{on}}\:\boldsymbol{{a}}\:\:\boldsymbol{{a}}_{\boldsymbol{{n}}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{........\left(\frac{\mathrm{1}}{\boldsymbol{{n}}+\mathrm{1}}\right)} } } .\:\:\boldsymbol{{a}}_{\mathrm{1}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:.\:\:\boldsymbol{{a}}_{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)} =\boldsymbol{{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{ln}}\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\:\:\boldsymbol{{a}}_{\mathrm{3}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{4}}\right)} } =\boldsymbol{{e}}^{\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{ln}}\left(\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)} \right)} =\boldsymbol{{e}}^{\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{ln}}\left(\boldsymbol{{a}}_{\mathrm{2}} \right)} =\boldsymbol{{e}}^{\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{ln}}\left(\boldsymbol{{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{ln}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)} =\boldsymbol{{e}}^{\frac{\mathrm{2}}{\mathrm{4}!}\boldsymbol{{ln}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\:\:\:\boldsymbol{{a}}_{\mathrm{4}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{5}}\right)} } } =\boldsymbol{{e}}^{\frac{\mathrm{1}}{\mathrm{5}}\boldsymbol{{ln}}\left(\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{4}}\right)} } \right)} =\boldsymbol{{e}}^{\frac{\mathrm{1}}{\mathrm{5}}\boldsymbol{{ln}}\left(\boldsymbol{{a}}_{\mathrm{3}} \right)} =\boldsymbol{{e}}^{\frac{\mathrm{1}}{\mathrm{5}}\boldsymbol{{ln}}\left(\boldsymbol{{e}}^{\frac{\mathrm{2}}{\mathrm{4}!}\boldsymbol{{ln}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)} =\boldsymbol{{e}}^{\frac{\mathrm{2}}{\mathrm{5}!}\boldsymbol{{ln}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:\boldsymbol{{a}}_{\mathrm{5}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{6}}\right)} } } } =\boldsymbol{{e}}^{\frac{\mathrm{1}}{\mathrm{6}}\boldsymbol{{ln}}\left(\boldsymbol{{a}}_{\mathrm{4}} \right)} =\boldsymbol{{e}}^{\frac{\mathrm{1}}{\mathrm{6}}\boldsymbol{{ln}}\left(\boldsymbol{{e}}^{\frac{\mathrm{2}}{\mathrm{5}!}\boldsymbol{{ln}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)} =\boldsymbol{{e}}^{\frac{\mathrm{2}}{\mathrm{6}!}\boldsymbol{{ln}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:.... \\ $$$$\:\:\:\:\:... \\ $$$$\:\:\:\boldsymbol{{a}}_{\boldsymbol{{n}}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{.....\left(\frac{\mathrm{1}}{\boldsymbol{{n}}+\mathrm{1}}\right)} } =\boldsymbol{{e}}^{\frac{\mathrm{1}}{\left(\boldsymbol{{n}}+\mathrm{1}\right)}\boldsymbol{{ln}}\left(\boldsymbol{{a}}_{\boldsymbol{{n}}−\mathrm{1}} \right)} =\boldsymbol{{e}}^{\frac{\mathrm{1}}{\left(\boldsymbol{{n}}+\mathrm{1}\right)}\boldsymbol{{ln}}\left(\boldsymbol{{e}}^{\frac{\mathrm{2}}{\boldsymbol{{n}}!}\boldsymbol{{ln}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)} =\boldsymbol{{e}}^{\frac{\mathrm{2}}{\left(\boldsymbol{{n}}+\mathrm{1}\right)!}\boldsymbol{{ln}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\boldsymbol{{et}}\:\::\boldsymbol{{n}}>\mathrm{0}\:\:\:\left(\boldsymbol{{n}}+\mathrm{1}\right)<\left(\boldsymbol{{n}}+\mathrm{1}\right)!<\left(\boldsymbol{{n}}+\mathrm{1}\right)^{\left(\boldsymbol{{n}}+\mathrm{1}\right)} \:\:\:\Rightarrow\:\:\:\boldsymbol{{n}}>\mathrm{0}\:\:\:\:\frac{\mathrm{2}}{\left(\boldsymbol{{n}}+\mathrm{1}\right)}>\frac{\mathrm{2}}{\left(\boldsymbol{{n}}+\mathrm{1}\right)!}>\frac{\mathrm{2}}{\left(\boldsymbol{{n}}+\mathrm{1}\right)^{\left(\boldsymbol{{n}}+\mathrm{1}\right)} }\: \\ $$$$\:\:\:\:\:\boldsymbol{{n}}>\mathrm{0}\:\:\:\:\:\:\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{2}}{\boldsymbol{{n}}+\mathrm{1}}\right)} <\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{2}}{\left(\boldsymbol{{n}}+\mathrm{1}\right)!}\right)} <\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\left(\boldsymbol{{n}}+\mathrm{1}\right)^{\left(\boldsymbol{{n}}+\mathrm{1}\right)} }} \:\:\Rightarrow\:\boldsymbol{{n}}>\mathrm{0}\:.\:\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{2}}{\boldsymbol{{n}}+\mathrm{1}}\right)} <\boldsymbol{{a}}_{\boldsymbol{{n}}} =\boldsymbol{{e}}^{\frac{\mathrm{2}}{\left(\boldsymbol{{n}}+\mathrm{1}\right)!}\boldsymbol{{ln}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} <\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\left(\boldsymbol{{n}}+\mathrm{1}\right)^{\left(\boldsymbol{{n}}+\mathrm{1}\right)} }} \\ $$$$\:\boldsymbol{{donc}}\:\:\::\:\boldsymbol{{li}}\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{m}}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\left(\boldsymbol{{n}}+\mathrm{1}\right)}} <\:\boldsymbol{{li}}\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{m}a}}_{\boldsymbol{{n}}} <\boldsymbol{{li}}\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{m}}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\left(\boldsymbol{{n}}+\mathrm{1}\right)^{\left(\boldsymbol{{n}}+\mathrm{1}\right)} }} \Rightarrow\:\:\:\boldsymbol{{n}}>\mathrm{0}\:\:\:\:\mathrm{1}<\boldsymbol{{li}}\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{m}a}}_{\boldsymbol{{n}}} <\mathrm{1} \\ $$$$\boldsymbol{{en}}\:\boldsymbol{{fin}}:\left(\:\boldsymbol{{li}}\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{m}a}}_{\boldsymbol{{n}}} =\mathrm{1}\right) \\ $$

Commented by Dwaipayan Shikari last updated on 24/Jun/20

I had a thought like you.Our thinking is matched. Great sir

$${I}\:\:{had}\:\:{a}\:\:{thought}\:{like}\:{you}.{Our}\:{thinking}\:{is}\:{matched}.\:{Great}\:{sir} \\ $$

Commented by Dwaipayan Shikari last updated on 24/Jun/20

Just matched!!But i was not so sure about answer. Sir you create my belief

$$\mathrm{Just}\:\mathrm{matched}!!\mathrm{But}\:\mathrm{i}\:\mathrm{was}\:\mathrm{not}\:\mathrm{so}\:\mathrm{sure}\:\mathrm{about}\:\mathrm{answer}.\:\mathrm{Sir}\:\mathrm{you}\:\mathrm{create}\:\mathrm{my}\:\mathrm{belief} \\ $$

Commented by bachamohamed last updated on 24/Jun/20

thank′s

$$\:{thank}'\mathrm{s} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com