Question Number 198152 by universe last updated on 11/Oct/23
$$\:\:\:\:{a}_{{n}+\mathrm{2}} \:=\:\:\:\sqrt{{a}_{{n}} ×{a}_{{n}+\mathrm{1}} }\:\:\:\forall\:{n}\geqslant\mathrm{1}\:,\:{n}\:\in\:\mathrm{N} \\ $$$$\:\mathrm{and}\:\mathrm{here}\:\:\mathrm{a}_{\mathrm{1}\:} =\:\alpha\:\:{and}\:{a}_{\mathrm{2}} =\:\beta\:\:\mathrm{then} \\ $$$$\:\:\:\mathrm{prove}\:\mathrm{that}\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}+\mathrm{2}} \:\:\:=\:\:\left(\alpha×\beta^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{3}} \\ $$
Answered by mr W last updated on 12/Oct/23
![let b_n =ln a_n a_(n+2) =(√(a_(n+1) ×a_n )) ln a_(n+2) =((ln a_(n+1) +ln a_n )/2) 2 ln a_(n+2) −ln a_(n+1) −ln a_n =0 ⇒2b_(n+2) −b_(n+1) −b_n =0 (recurrence relation) 2p^2 −p−1=0 (characteristic equation) (2p+1)(p−1)=0 ⇒p=1, −(1/2) ⇒b_n =A+B(−(1/2))^n b_1 =A+B(−(1/2))=ln a_1 =ln α ...(i) b_2 =A+B(−(1/2))^2 =ln a_2 =ln β ...(ii) (ii)−(i): ((3B)/4)=ln β−ln α ⇒B=(4/3)(ln (β/α))=ln ((β/α))^(4/3) ⇒A=(B/2)+ln α=(2/3)(ln (β/α))+ln α=ln (αβ^2 )^(1/3) ⇒b_n =ln (αβ^2 )^(1/3) +(−(1/2))^n ln ((β/α))^(4/3) ⇒b_n =ln [(αβ^2 )^(1/3) ((β/α))^((4/3)(−(1/2))^n ) ]=ln a_n ⇒a_n = (αβ^2 )^(1/3) ((β/α))^((4/3)(−(1/2))^n ) ⇒lim_(n→∞) a_n =(αβ^2 )^(1/3) ✓](https://www.tinkutara.com/question/Q198157.png)
$${let}\:{b}_{{n}} =\mathrm{ln}\:{a}_{{n}} \\ $$$${a}_{{n}+\mathrm{2}} =\sqrt{{a}_{{n}+\mathrm{1}} ×{a}_{{n}} } \\ $$$$\mathrm{ln}\:{a}_{{n}+\mathrm{2}} =\frac{\mathrm{ln}\:{a}_{{n}+\mathrm{1}} +\mathrm{ln}\:{a}_{{n}} }{\mathrm{2}} \\ $$$$\mathrm{2}\:\mathrm{ln}\:{a}_{{n}+\mathrm{2}} −\mathrm{ln}\:{a}_{{n}+\mathrm{1}} −\mathrm{ln}\:{a}_{{n}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{b}_{{n}+\mathrm{2}} −{b}_{{n}+\mathrm{1}} −{b}_{{n}} =\mathrm{0}\:\:\:\left({recurrence}\:{relation}\right) \\ $$$$\mathrm{2}{p}^{\mathrm{2}} −{p}−\mathrm{1}=\mathrm{0}\:\:\:\:\left({characteristic}\:{equation}\right) \\ $$$$\left(\mathrm{2}{p}+\mathrm{1}\right)\left({p}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{p}=\mathrm{1},\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{b}_{{n}} ={A}+{B}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$${b}_{\mathrm{1}} ={A}+{B}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{ln}\:{a}_{\mathrm{1}} =\mathrm{ln}\:\alpha\:\:\:…\left({i}\right) \\ $$$${b}_{\mathrm{2}} ={A}+{B}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{ln}\:{a}_{\mathrm{2}} =\mathrm{ln}\:\beta\:\:\:…\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$$\frac{\mathrm{3}{B}}{\mathrm{4}}=\mathrm{ln}\:\beta−\mathrm{ln}\:\alpha \\ $$$$\Rightarrow{B}=\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{ln}\:\frac{\beta}{\alpha}\right)=\mathrm{ln}\:\left(\frac{\beta}{\alpha}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} \\ $$$$\Rightarrow{A}=\frac{{B}}{\mathrm{2}}+\mathrm{ln}\:\alpha=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{ln}\:\frac{\beta}{\alpha}\right)+\mathrm{ln}\:\alpha=\mathrm{ln}\:\left(\alpha\beta^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\Rightarrow{b}_{{n}} =\mathrm{ln}\:\left(\alpha\beta^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \mathrm{ln}\:\left(\frac{\beta}{\alpha}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} \\ $$$$\Rightarrow{b}_{{n}} =\mathrm{ln}\:\left[\left(\alpha\beta^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\frac{\beta}{\alpha}\right)^{\frac{\mathrm{4}}{\mathrm{3}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} } \right]=\mathrm{ln}\:{a}_{{n}} \\ $$$$\Rightarrow{a}_{{n}} =\:\left(\alpha\beta^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\frac{\beta}{\alpha}\right)^{\frac{\mathrm{4}}{\mathrm{3}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} } \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} =\left(\alpha\beta^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\checkmark \\ $$
Commented by universe last updated on 12/Oct/23
$$\:\:{thanks}\:{sir} \\ $$
Answered by Frix last updated on 12/Oct/23
$${a}_{{n}+\mathrm{2}} =\sqrt{{a}_{{n}+\mathrm{1}} {a}_{{n}} } \\ $$$${a}_{{n}} =\mathrm{e}^{{C}_{\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} +{C}_{\mathrm{2}} } \\ $$$${a}_{\mathrm{1}} =\alpha\:\Leftrightarrow\:\mathrm{e}^{−\frac{{C}_{\mathrm{1}} }{\mathrm{2}}+{C}_{\mathrm{2}} } =\alpha\:\Leftrightarrow\:−\frac{{C}_{\mathrm{1}} }{\mathrm{2}}+{C}_{\mathrm{2}} =\mathrm{ln}\:\alpha \\ $$$${a}_{\mathrm{2}} =\beta\:\Leftrightarrow\:\mathrm{e}^{\frac{{C}_{\mathrm{1}} }{\mathrm{4}}+{C}_{\mathrm{2}} } =\beta\:\Leftrightarrow\:\frac{{C}_{\mathrm{1}} }{\mathrm{4}}+{C}_{\mathrm{2}} =\mathrm{ln}\:\beta \\ $$$$\Rightarrow \\ $$$${C}_{\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{3}}\mathrm{ln}\:\frac{\beta}{\alpha}\:\wedge{C}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\alpha\beta^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${a}_{{n}} =\alpha^{\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\left(−\mathrm{2}\right)^{\mathrm{2}−{n}} \right)} \beta^{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}+\left(−\mathrm{2}\right)^{\mathrm{1}−{n}} \right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} \:=\alpha^{\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}\pm\mathrm{0}\right)} \beta^{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}\mp\mathrm{0}\right)} =\left(\alpha\beta^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$