Question Number 223778 by wewji12 last updated on 04/Aug/25
$$\underset{{N}\rightarrow\infty} {\mathrm{lim}}\:\frac{{p}_{{N}+\mathrm{1}} }{{p}_{{N}} }=???\:\:,\:{p}_{\mathrm{1}} =\mathrm{2}\:,\:{p}_{\mathrm{2}} =\mathrm{3}\:,{p}_{\mathrm{3}} =\mathrm{5}\:…. \\ $$
Answered by mr W last updated on 04/Aug/25
$${p}_{{n}+\mathrm{1}} ={p}_{{n}} +{p}_{{n}−\mathrm{1}} \\ $$$$\frac{{p}_{{n}+\mathrm{1}} }{{p}^{{n}} }=\mathrm{1}+\frac{{p}_{{n}−\mathrm{1}} }{{p}_{{n}} } \\ $$$${say}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{p}_{{n}+\mathrm{1}} }{{p}_{{n}} }=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{p}_{{n}} }{{p}_{{n}−\mathrm{1}} }={L}\:>\mathrm{1} \\ $$$${L}=\mathrm{1}+\frac{\mathrm{1}}{{L}} \\ $$$${L}^{\mathrm{2}} −{L}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{L}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\varphi\approx\mathrm{1}.\mathrm{618}\:\:\:\:\:\:\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}<\mathrm{1}\:\Rightarrow{rejected}\right) \\ $$
Commented by Raphael254 last updated on 04/Aug/25
$${but}\:{p}_{{n}} \:{is}\:{not}\:{a}\:{prime}\:{number}? \\ $$$${i}\:{think}\:{p}_{\mathrm{1}} \:=\:\mathrm{2},\:{p}_{\mathrm{2}} \:=\:\mathrm{3},\:{p}_{\mathrm{3}} \:=\:\mathrm{5},\:{means}\:{the}\:{first}\:\mathrm{3}\:{prime}\:{numbers},\:{can}\:{also}\:{means}\:{the}\:{fourth},\:{fifth}\:{and}\:{sixth}\:{terms}\:{of}\:{Fibonacci}'{s}\:{sequence}\:\left(\mathrm{0},\:\mathrm{1},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{5}\right). \\ $$$${But}\:{i}\:{don}'{t}\:{know} \\ $$
Commented by mr W last updated on 04/Aug/25
$${then}\:{it}\:{should}\:{be}\:{said}\:{that}\:{p}_{{i}} \:{are} \\ $$$${prime}\:{numbers}!\:{who}\:{can}\:{otherwise} \\ $$$${know}\:{what}\:{is}\:{really}\:{meant}?\: \\ $$$${i}\:{thought}\:{it}\:{meant}\:\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8},\mathrm{13},\mathrm{21},… \\ $$
Commented by Raphael254 last updated on 04/Aug/25
$${the}\:{p}_{\mathrm{1}} =\:\mathrm{2},\:{p}_{\mathrm{2}} \:=\:\mathrm{3},\:{p}_{\mathrm{3}} \:=\:\mathrm{5}\:{of}\:{the}\:{question},\:{it}\:{could}\:{be}\:{the}\:{fourth},\:{fifth}\:{and}\:{sixth}\:{terms}\:{of}\:{the}\:{Fibonacci}'{s}\:{sequence}, \\ $$$${this}\:{question}\:{is}\:{a}\:{little}\:{ambiguous},\:{in}\:{a}\:{common}\:{sense},\:{we}\:{know}\:{that}\:{the}\:{first}\:{three}\:{prime}\:{numbers}\:{are}\:\mathrm{2},\:\mathrm{3}\:{and}\:\mathrm{5},\:{so}\:{p}_{\mathrm{1}} ,\:{p}_{\mathrm{2}} \:{and}\:{p}_{\mathrm{3}} \:{could}\:{be}\:{prime}\:{numbers}. \\ $$$${It}\:{can}\:{also}\:{be}\:{the}\:{Fibonacci}'{s}\:{sequence},\:{but}\:{started}\:{at}\:{the}\:{fourth}\:{term}. \\ $$$$ \\ $$$${This}\:{ambiguity}\:{could}\:{be}\:{easily}\:{broken}\:{if}\:{the}\:{sequence}\:{was}\:{or}\:{was}\:{not}:\:{p}_{\mathrm{1}} \:=\:\mathrm{2},\:{p}_{\mathrm{2}} \:=\:\mathrm{3},\:{p}_{\mathrm{3}} \:=\:\mathrm{5},\:{p}_{\mathrm{4}} \:=\:\mathrm{8}\:… \\ $$$$ \\ $$$${But}\:{another}\:{thing}\:{that}\:{tells}\:{that}\:{the}\:{sequence}\:{is}\:{formed}\:{by}\:{prime}\:{numbers}\:{is}\:{the}\:\boldsymbol{{p}}\:{possibly}\:{making}\:{reference}\:{to}\:\boldsymbol{{prime}}, \\ $$$${but}\:{i}\:{don}'{t}\:{know}.\:{I}\:{only}\:{replied}\:{because}\:{i}\:{understood}\:{it}\:{in}\:{a}\:{different}\:{way}. \\ $$
Commented by Frix last updated on 04/Aug/25
$$\mathrm{It}'\mathrm{s}\:\mathrm{three}\:\mathrm{given}\:\mathrm{numbers},\:\mathrm{for}\:\mathrm{me}\:“{p}_{{n}} ''\:\mathrm{means} \\ $$$$“\mathrm{polynomial}'' \\ $$$$\Rightarrow \\ $$$${p}_{{n}} =\frac{{n}^{\mathrm{2}} }{\mathrm{2}}−\frac{{n}}{\mathrm{2}}+\mathrm{2} \\ $$$$\Rightarrow\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{p}_{{n}+\mathrm{1}} }{{p}_{{n}} }\:=\mathrm{1} \\ $$