Question Number 2297 by prakash jain last updated on 14/Nov/15
$${a}_{\mathrm{0}} ={k} \\ $$$${a}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}+{a}_{{n}} } \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} =? \\ $$
Commented by Yozzi last updated on 15/Nov/15
![a_1 =(1/(1+a_0 ))=(1/(1+k))=(((0)k+1)/(k+1)) a_2 =(1/(1+a_1 ))=(1/(1+(1/(1+k))))=((k+1)/(k+2)) a_3 =(1/(1+a_2 ))=(1/(1+((k+1)/(k+2))))=((k+2)/(2k+3)) a_4 =(1/(1+a_3 ))=(1/(1+((k+2)/(2k+3))))=((2k+3)/(3k+5)) a_5 =(1/(1+a_4 ))=(1/(1+((2k+3)/(3k+5))))=((3k+5)/(5k+8)) a_6 =(1/(1+a_5 ))=(1/(1+((3k+5)/(5k+8))))=((5k+8)/(8k+13)) a_n =((F_(n−1) k+F_n )/(F_n k+F_(n+1) )), n≥1 F_i =(1/(√5))[(((1+(√5))/2))^i −(((1−(√5))/2))^i ],i≥0 Let r_i (+)=(((1+(√5))/2))^i ,r_i (−)=(((1−(√5))/2))^i ∴F_i =(1/(√5))(r_i (+)−r_i (−)) ∴ a_n =(((1/(√5))(k[r_(n−1) (+)−r_(n−1) (−)]+r_n (+)−r_n (−)))/((1/(√5))(k[r_n (+)−r_n (−)]+r_(n+1) (+)−r_(n+1) (−)))) a_n =((k(r_(n−1) (+)−r_(n−1) (−))+r_n (+)−r_n (−))/(k(r_n (+)−r_n (−))+r_(n+1) (+)−r_(n+1) (−))) a_n =((k((1/(r_2 (+)))−((r_(n−1) (−))/(r_(n+1) (+))))+(1/(r_1 (+)))−((r_n (−))/(r_(n+1) (+))))/(k((1/(r_1 (+)))−((r_n (−))/(r_(n+1) (+))))+1−((r_(n+1) (−))/(r_(n+1) (+))))) ((r_n (−))/(r_(n+1) (+)))=(((((1−(√5))/2))^n )/((((1+(√5))/2))^(n+1) ))=(((1−(√5))/(1+(√5))))^n ×(2/(1+(√5))) ∣((1−(√5))/(1+(√5)))∣<1⇒(((1−(√5))/(1+(√5))))^n →0 as n→∞ ⇒lim_(n→∞) ((r_n (−))/(r_(n+1) (+)))=0 ((r_(n−1) (−))/(r_(n+1) (+)))=(((1−(√5))/(1+(√5))))^n ×(4/((1+(√5))(1−(√5)))) ∴lim_(n→∞) ((r_(n−1) (−))/(r_(n+1) (+)))=0 ((r_(n+1) (−))/(r_(n+1) (+)))=(((1−(√5))/(1+(√5))))^(n+1) ∴lim_(n→∞) ((r_(n+1) (−))/(r_(n+1) (+)))=0 ∴ lim_(n→∞) a_n =(((k/(r_2 (+)))+(1/(r_1 (+))))/((k/(r_1 (+)))+1))=((kr_1 (+)+r_2 (+))/(r_2 (+)(k+r_1 (+)))) (Comment the mistake please)](Q2303.png)
$$ \\ $$$${a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}+{a}_{\mathrm{0}} }=\frac{\mathrm{1}}{\mathrm{1}+{k}}=\frac{\left(\mathrm{0}\right){k}+\mathrm{1}}{{k}+\mathrm{1}} \\ $$$${a}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}+{a}_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+{k}}}=\frac{{k}+\mathrm{1}}{{k}+\mathrm{2}} \\ $$$${a}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{1}+{a}_{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{1}+\frac{{k}+\mathrm{1}}{{k}+\mathrm{2}}}=\frac{{k}+\mathrm{2}}{\mathrm{2}{k}+\mathrm{3}} \\ $$$${a}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{1}+{a}_{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{1}+\frac{{k}+\mathrm{2}}{\mathrm{2}{k}+\mathrm{3}}}=\frac{\mathrm{2}{k}+\mathrm{3}}{\mathrm{3}{k}+\mathrm{5}} \\ $$$${a}_{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{1}+{a}_{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}{k}+\mathrm{3}}{\mathrm{3}{k}+\mathrm{5}}}=\frac{\mathrm{3}{k}+\mathrm{5}}{\mathrm{5}{k}+\mathrm{8}} \\ $$$${a}_{\mathrm{6}} =\frac{\mathrm{1}}{\mathrm{1}+{a}_{\mathrm{5}} }=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{3}{k}+\mathrm{5}}{\mathrm{5}{k}+\mathrm{8}}}=\frac{\mathrm{5}{k}+\mathrm{8}}{\mathrm{8}{k}+\mathrm{13}} \\ $$$${a}_{{n}} =\frac{{F}_{{n}−\mathrm{1}} {k}+{F}_{{n}} }{{F}_{{n}} {k}+{F}_{{n}+\mathrm{1}} },\:{n}\geqslant\mathrm{1} \\ $$$${F}_{{i}} =\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{i}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{i}} \right],{i}\geqslant\mathrm{0} \\ $$$${Let}\:{r}_{{i}} \left(+\right)=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{i}} ,{r}_{{i}} \left(−\right)=\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{i}} \\ $$$$\therefore{F}_{{i}} =\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left({r}_{{i}} \left(+\right)−{r}_{{i}} \left(−\right)\right) \\ $$$$\therefore\:{a}_{{n}} =\frac{\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left({k}\left[{r}_{{n}−\mathrm{1}} \left(+\right)−{r}_{{n}−\mathrm{1}} \left(−\right)\right]+{r}_{{n}} \left(+\right)−{r}_{{n}} \left(−\right)\right)}{\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left({k}\left[{r}_{{n}} \left(+\right)−{r}_{{n}} \left(−\right)\right]+{r}_{{n}+\mathrm{1}} \left(+\right)−{r}_{{n}+\mathrm{1}} \left(−\right)\right)} \\ $$$${a}_{{n}} =\frac{{k}\left({r}_{{n}−\mathrm{1}} \left(+\right)−{r}_{{n}−\mathrm{1}} \left(−\right)\right)+{r}_{{n}} \left(+\right)−{r}_{{n}} \left(−\right)}{{k}\left({r}_{{n}} \left(+\right)−{r}_{{n}} \left(−\right)\right)+{r}_{{n}+\mathrm{1}} \left(+\right)−{r}_{{n}+\mathrm{1}} \left(−\right)} \\ $$$${a}_{{n}} =\frac{{k}\left(\frac{\mathrm{1}}{{r}_{\mathrm{2}} \left(+\right)}−\frac{{r}_{{n}−\mathrm{1}} \left(−\right)}{{r}_{{n}+\mathrm{1}} \left(+\right)}\right)+\frac{\mathrm{1}}{{r}_{\mathrm{1}} \left(+\right)}−\frac{{r}_{{n}} \left(−\right)}{{r}_{{n}+\mathrm{1}} \left(+\right)}}{{k}\left(\frac{\mathrm{1}}{{r}_{\mathrm{1}} \left(+\right)}−\frac{{r}_{{n}} \left(−\right)}{{r}_{{n}+\mathrm{1}} \left(+\right)}\right)+\mathrm{1}−\frac{{r}_{{n}+\mathrm{1}} \left(−\right)}{{r}_{{n}+\mathrm{1}} \left(+\right)}} \\ $$$$\frac{{r}_{{n}} \left(−\right)}{{r}_{{n}+\mathrm{1}} \left(+\right)}=\frac{\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} }{\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}+\mathrm{1}} }=\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\right)^{{n}} ×\frac{\mathrm{2}}{\mathrm{1}+\sqrt{\mathrm{5}}} \\ $$$$\mid\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\mid<\mathrm{1}\Rightarrow\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\right)^{{n}} \rightarrow\mathrm{0}\:{as}\:{n}\rightarrow\infty \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{r}_{{n}} \left(−\right)}{{r}_{{n}+\mathrm{1}} \left(+\right)}=\mathrm{0} \\ $$$$\frac{{r}_{{n}−\mathrm{1}} \left(−\right)}{{r}_{{n}+\mathrm{1}} \left(+\right)}=\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\right)^{{n}} ×\frac{\mathrm{4}}{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)} \\ $$$$\therefore\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{r}_{{n}−\mathrm{1}} \left(−\right)}{{r}_{{n}+\mathrm{1}} \left(+\right)}=\mathrm{0} \\ $$$$\frac{{r}_{{n}+\mathrm{1}} \left(−\right)}{{r}_{{n}+\mathrm{1}} \left(+\right)}=\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\right)^{{n}+\mathrm{1}} \\ $$$$\therefore\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{r}_{{n}+\mathrm{1}} \left(−\right)}{{r}_{{n}+\mathrm{1}} \left(+\right)}=\mathrm{0} \\ $$$$\therefore\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} =\frac{\frac{{k}}{{r}_{\mathrm{2}} \left(+\right)}+\frac{\mathrm{1}}{{r}_{\mathrm{1}} \left(+\right)}}{\frac{{k}}{{r}_{\mathrm{1}} \left(+\right)}+\mathrm{1}}=\frac{{kr}_{\mathrm{1}} \left(+\right)+{r}_{\mathrm{2}} \left(+\right)}{{r}_{\mathrm{2}} \left(+\right)\left({k}+{r}_{\mathrm{1}} \left(+\right)\right)} \\ $$$$\left({Comment}\:{the}\:{mistake}\:{please}\right) \\ $$$$ \\ $$
Commented by Yozzi last updated on 14/Nov/15
$${Sorry}.\:{Mistake}. \\ $$
Commented by prakash jain last updated on 14/Nov/15
$$\mathrm{If}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{is}\:\mathrm{such}\:\mathrm{that}\:{a}_{{n}} \neq−\mathrm{1},\:\:\forall{n}\in\mathbb{N} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{always}\:\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Yozzi last updated on 14/Nov/15
$${I}\:{see}.\:{Nice}! \\ $$
Answered by prakash jain last updated on 14/Nov/15
$$\mathrm{If}\:\mathrm{the}\:\mathrm{limits}\:\mathrm{exists}\:\left(\mathrm{when}\:{a}_{{n}} \neq−\mathrm{1}\forall{n}\in\mathbb{N}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}+\mathrm{1}} ={a}_{{n}} \\ $$$${a}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}+{a}_{{n}} }\Rightarrow{a}_{{n}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$