Question Number 78156 by Rio Michael last updated on 14/Jan/20
$$\mathrm{Given}\:\mathrm{that}\:{u}_{{n}\:+\:\mathrm{1}} =\:\frac{{a}_{{n}} }{\mathrm{2}}\:+\:\mathrm{5}\:\:\: \\ $$$${evalatuate}\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} \\ $$$${deduce}\:{if}\:{a}_{{n}\:} \:{is}\:{convergent}\:{or}\:{divergent}. \\ $$
Commented by mathmax by abdo last updated on 14/Jan/20
$$\Rightarrow\mathrm{2}{u}_{{n}+\mathrm{1}} ={u}_{{n}} +\mathrm{10}\:\Rightarrow\mathrm{2}{u}_{{n}+\mathrm{1}} −{u}_{{n}} −\mathrm{10}=\mathrm{0} \\ $$$$\left({ce}\right)\rightarrow\mathrm{2}{r}−\mathrm{1}=\mathrm{0}\:\Rightarrow{r}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{u}_{{n}} =\frac{\lambda}{\mathrm{2}^{{n}} }+\beta \\ $$$${u}_{\mathrm{0}} =\lambda+\beta \\ $$$${u}_{\mathrm{1}} =\frac{{u}_{\mathrm{0}} }{\mathrm{2}}+\mathrm{5}\:=\frac{\lambda}{\mathrm{2}}\:+\beta\:\Rightarrow{u}_{\mathrm{0}} \:+\mathrm{10}\:=\lambda+\mathrm{2}\beta\:\Rightarrow{u}_{\mathrm{0}} \:+\mathrm{10}\:=\lambda+\mathrm{2}\left({u}_{\mathrm{0}} −\lambda\right)\:\Rightarrow \\ $$$${u}_{\mathrm{0}} \:+\mathrm{10}\:=−\lambda+\mathrm{2}{u}_{\mathrm{0}} \:\Rightarrow\lambda\:=\mathrm{2}{u}_{\mathrm{0}} −{u}_{\mathrm{0}} −\mathrm{10}\:={u}_{\mathrm{0}} −\mathrm{10} \\ $$$$\beta\:={u}_{\mathrm{0}} −\lambda\:={u}_{\mathrm{0}} −{u}_{\mathrm{0}} \:+\mathrm{10}\:=\mathrm{10}\:\Rightarrow{u}_{{n}} =\frac{{u}_{\mathrm{0}} −\mathrm{10}}{\mathrm{2}^{{n}} }\:+\mathrm{10} \\ $$$${if}\:{u}_{\mathrm{0}} \neq\mathrm{10}\:\:\:{lim}_{{n}\rightarrow+\infty} \:{u}_{{n}} =\mathrm{10} \\ $$$${if}\:{u}_{\mathrm{0}} =\mathrm{10}\:\:\Rightarrow{u}_{{n}} =\mathrm{10}\:\forall{n}\:\Rightarrow{lim}\:{u}_{{n}} =\mathrm{10} \\ $$$$ \\ $$