Question Number 88471 by M±th+et£s last updated on 10/Apr/20
Answered by mr W last updated on 10/Apr/20
$${a}_{{n}+\mathrm{1}} =\mathrm{2}+\frac{\mathrm{5}}{{a}_{{n}} } \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}+\mathrm{1}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}+\frac{\mathrm{5}}{{a}_{{n}} }\right) \\ $$$${l}=\mathrm{2}+\frac{\mathrm{5}}{{l}} \\ $$$${l}^{\mathrm{2}} −\mathrm{2}{l}−\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow{l}=\mathrm{1}+\sqrt{\mathrm{6}} \\ $$
Commented by M±th+et£s last updated on 10/Apr/20
$${thanx}\:{sir}\:{but}\:{why}\:\underset{{n}\rightarrow\infty} {{lim}a}_{{n}+\mathrm{1}} ={a}_{{n}} \\ $$
Commented by M±th+et£s last updated on 10/Apr/20
$${sorry}\:{i}\:{mean}\:\underset{{n}\rightarrow\infty} {{lim}}\:{a}_{{n}} \\ $$
Commented by M±th+et£s last updated on 11/Apr/20
$${and}\:{can}\:{we}\:{find}\:{a}_{{n}} \\ $$
Commented by mr W last updated on 11/Apr/20
$${if}\:{convergence}\:{exists},\:{i}.{e}. \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} ={l},\:{that}\:{means}\:{when}\:{n}\rightarrow\infty, \\ $$$${a}_{{n}} \rightarrow{l} \\ $$$${then}\:{it}'{s}\:{clear},\:{when}\:{n}\rightarrow\infty,\:\Rightarrow{n}+\mathrm{2}\rightarrow\infty \\ $$$$\Rightarrow{a}_{{n}+\mathrm{2}} \rightarrow{l},\:{therefore}: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}+\mathrm{2}} ={l} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}+\mathrm{2000000}} ={l} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}−\mathrm{100}} ={l} \\ $$
Commented by mr W last updated on 11/Apr/20
$${i}'{m}\:{not}\:{sure}\:{that}\:{we}\:{can}\:{get}\:{the}\:{explicit} \\ $$$${form}\:{for}\:{a}_{{n}} . \\ $$
Commented by M±th+et£s last updated on 11/Apr/20
$${thank}\:{you}\:{sir} \\ $$