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let-u-0-1-and-u-n-1-u-n-2-u-n-study-the-convervence-of-u-n-




Question Number 42789 by maxmathsup by imad last updated on 02/Sep/18
let u_0 =1 and u_(n+1)  =u_n  + (2/u_n )  study the convervence of (u_n )
$${let}\:{u}_{\mathrm{0}} =\mathrm{1}\:{and}\:{u}_{{n}+\mathrm{1}} \:={u}_{{n}} \:+\:\frac{\mathrm{2}}{{u}_{{n}} } \\ $$$${study}\:{the}\:{convervence}\:{of}\:\left({u}_{{n}} \right) \\ $$
Commented by maxmathsup by imad last updated on 30/Sep/18
let prove that u_n >0   u_0 =1>0 letsuppse u_n >0 ⇒u_n  +(2/u_n )>0 ⇒u_(n+1) >0  we have u_(n+1) −u_n =(2/u_n )>0 ⇒ (u_n ) is increasing  if  (u_n ) converges to a number l we get  l=f(l) with f(x)=x +(2/x) ⇒  l=l+(2/l) ⇒(2/l) =0 and this is impossible so  (u_n ) is not convergent .
$${let}\:{prove}\:{that}\:{u}_{{n}} >\mathrm{0}\:\:\:{u}_{\mathrm{0}} =\mathrm{1}>\mathrm{0}\:{letsuppse}\:{u}_{{n}} >\mathrm{0}\:\Rightarrow{u}_{{n}} \:+\frac{\mathrm{2}}{{u}_{{n}} }>\mathrm{0}\:\Rightarrow{u}_{{n}+\mathrm{1}} >\mathrm{0} \\ $$$${we}\:{have}\:{u}_{{n}+\mathrm{1}} −{u}_{{n}} =\frac{\mathrm{2}}{{u}_{{n}} }>\mathrm{0}\:\Rightarrow\:\left({u}_{{n}} \right)\:{is}\:{increasing} \\ $$$${if}\:\:\left({u}_{{n}} \right)\:{converges}\:{to}\:{a}\:{number}\:{l}\:{we}\:{get}\:\:{l}={f}\left({l}\right)\:{with}\:{f}\left({x}\right)={x}\:+\frac{\mathrm{2}}{{x}}\:\Rightarrow \\ $$$${l}={l}+\frac{\mathrm{2}}{{l}}\:\Rightarrow\frac{\mathrm{2}}{{l}}\:=\mathrm{0}\:{and}\:{this}\:{is}\:{impossible}\:{so}\:\:\left({u}_{{n}} \right)\:{is}\:{not}\:{convergent}\:. \\ $$$$ \\ $$

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