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Question-204334




Question Number 204334 by universe last updated on 13/Feb/24
Answered by Frix last updated on 13/Feb/24
n→∞ ⇒ a_n =(√(2a_n )) ⇒ a_n =2
$${n}\rightarrow\infty\:\Rightarrow\:{a}_{{n}} =\sqrt{\mathrm{2}{a}_{{n}} }\:\Rightarrow\:{a}_{{n}} =\mathrm{2} \\ $$
Answered by MM42 last updated on 14/Feb/24
step 1:  a_(n+2) ^2 −a_(n+1) ^2 =a_(n+1) −a_(n−1) =((a_(n+1) ^2 −a_(n−1) ^2 )/(a_(n+1) +a_(n−1) ))=((a_(n−1) −a_(n−2) )/(a_(n+1) +a_(n−1) ))  =((a_(n−1) ^2 −a_(n−2) ^2 )/((a_(b+1) +a_(n−1) )(a_(n−1) +a_(n−2) )))  =((a_(n−2) −a_(n−3) )/((a_(n+1) +a_(n−1) )(a_(n−1) +a_(n−2) )))=...  =((a_2 −a_1 )/((a_(n+1) +a_(n−1) )(a_(n−1) +a_(n−2) )...(a_3 +a_2 )))>0  ⇒{a_n } ↗  step 2:  a_1 =0<2  , a_2 =1<2  , a_3 =1<2  a_4 =(√2)<2   ,  a_5 =(√(2+(√2)))<2  ∀ m<n→a_m <2  a_n ^2 =a_(n−2) +a_(n−1) <4 ⇒a_n <2 ✓  ⇒{a_n }  is convergent  let  a_n →a  ⇒lim_(n→∞) a_(n+2) =lim_(n→∞) (√(a_n +a_(n+1) ))  ⇒a=(√(2a))⇒ a=2  ✓
$${step}\:\mathrm{1}: \\ $$$${a}_{{n}+\mathrm{2}} ^{\mathrm{2}} −{a}_{{n}+\mathrm{1}} ^{\mathrm{2}} ={a}_{{n}+\mathrm{1}} −{a}_{{n}−\mathrm{1}} =\frac{{a}_{{n}+\mathrm{1}} ^{\mathrm{2}} −{a}_{{n}−\mathrm{1}} ^{\mathrm{2}} }{{a}_{{n}+\mathrm{1}} +{a}_{{n}−\mathrm{1}} }=\frac{{a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} }{{a}_{{n}+\mathrm{1}} +{a}_{{n}−\mathrm{1}} } \\ $$$$=\frac{{a}_{{n}−\mathrm{1}} ^{\mathrm{2}} −{a}_{{n}−\mathrm{2}} ^{\mathrm{2}} }{\left({a}_{{b}+\mathrm{1}} +{a}_{{n}−\mathrm{1}} \right)\left({a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} \right)} \\ $$$$=\frac{{a}_{{n}−\mathrm{2}} −{a}_{{n}−\mathrm{3}} }{\left({a}_{{n}+\mathrm{1}} +{a}_{{n}−\mathrm{1}} \right)\left({a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} \right)}=… \\ $$$$=\frac{{a}_{\mathrm{2}} −{a}_{\mathrm{1}} }{\left({a}_{{n}+\mathrm{1}} +{a}_{{n}−\mathrm{1}} \right)\left({a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} \right)…\left({a}_{\mathrm{3}} +{a}_{\mathrm{2}} \right)}>\mathrm{0} \\ $$$$\Rightarrow\left\{{a}_{{n}} \right\}\:\nearrow \\ $$$${step}\:\mathrm{2}: \\ $$$${a}_{\mathrm{1}} =\mathrm{0}<\mathrm{2}\:\:,\:{a}_{\mathrm{2}} =\mathrm{1}<\mathrm{2}\:\:,\:{a}_{\mathrm{3}} =\mathrm{1}<\mathrm{2} \\ $$$${a}_{\mathrm{4}} =\sqrt{\mathrm{2}}<\mathrm{2}\:\:\:,\:\:{a}_{\mathrm{5}} =\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}<\mathrm{2} \\ $$$$\forall\:{m}<{n}\rightarrow{a}_{{m}} <\mathrm{2} \\ $$$${a}_{{n}} ^{\mathrm{2}} ={a}_{{n}−\mathrm{2}} +{a}_{{n}−\mathrm{1}} <\mathrm{4}\:\Rightarrow{a}_{{n}} <\mathrm{2}\:\checkmark \\ $$$$\Rightarrow\left\{{a}_{{n}} \right\}\:\:{is}\:{convergent} \\ $$$${let}\:\:{a}_{{n}} \rightarrow{a} \\ $$$$\Rightarrow{lim}_{{n}\rightarrow\infty} {a}_{{n}+\mathrm{2}} ={lim}_{{n}\rightarrow\infty} \sqrt{{a}_{{n}} +{a}_{{n}+\mathrm{1}} } \\ $$$$\Rightarrow{a}=\sqrt{\mathrm{2}{a}}\Rightarrow\:{a}=\mathrm{2}\:\:\checkmark \\ $$$$ \\ $$

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