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Question Number 4589 by FilupSmith last updated on 09/Feb/16

a_n =(√(2+a_(n−1) ))  a_1 =(√2)    L=lim_(n→∞)  a_n =(√(2+(√(2+(√(2+(√(2+...))))))))  L=?

$${a}_{{n}} =\sqrt{\mathrm{2}+{a}_{{n}−\mathrm{1}} } \\ $$$${a}_{\mathrm{1}} =\sqrt{\mathrm{2}} \\ $$$$ \\ $$$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} =\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+...}}}} \\ $$$${L}=? \\ $$

Answered by Rasheed Soomro last updated on 09/Feb/16

a_n =(√(2+a_(n−1) ))  a_1 =2  a_2 =(√(2+a_1 ))=(√(2+2))=2  a_3 =(√(2+a_2 ))=(√(2+2))=2  This suggests that  a_n =2  How a_n =(√(2+(√(2+(√(2+(√(2+...))))))))   ?  ....................................................  Answer for corrected question  a_n =(√(2+a_(n−1) ))  a_1 =(√2)  L=(√(2+(√(2+(√(2+(√(2+...))))))))   L^2 =2+(√(2+(√(2+(√(2+(√(2+...))))))))   L^2 =2+L  L^2 −L−2=0  (L−2)(L+1)=0  L=2  ∣  L=−1  (√(2+(√(2+(√(2+(√(2+...))))))))   may not be negative  and  there is also chance of extraneous  roots, so I think −1 is extraneous root.  So L=2

$${a}_{{n}} =\sqrt{\mathrm{2}+{a}_{{n}−\mathrm{1}} } \\ $$$${a}_{\mathrm{1}} =\mathrm{2} \\ $$$${a}_{\mathrm{2}} =\sqrt{\mathrm{2}+{a}_{\mathrm{1}} }=\sqrt{\mathrm{2}+\mathrm{2}}=\mathrm{2} \\ $$$${a}_{\mathrm{3}} =\sqrt{\mathrm{2}+{a}_{\mathrm{2}} }=\sqrt{\mathrm{2}+\mathrm{2}}=\mathrm{2} \\ $$$${This}\:{suggests}\:{that} \\ $$$${a}_{{n}} =\mathrm{2} \\ $$$${How}\:{a}_{{n}} =\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+...}}}}\:\:\:? \\ $$$$.................................................... \\ $$$${Answer}\:{for}\:{corrected}\:{question} \\ $$$${a}_{{n}} =\sqrt{\mathrm{2}+{a}_{{n}−\mathrm{1}} } \\ $$$${a}_{\mathrm{1}} =\sqrt{\mathrm{2}} \\ $$$${L}=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+...}}}}\: \\ $$$${L}^{\mathrm{2}} =\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+...}}}}\: \\ $$$${L}^{\mathrm{2}} =\mathrm{2}+{L} \\ $$$${L}^{\mathrm{2}} −{L}−\mathrm{2}=\mathrm{0} \\ $$$$\left({L}−\mathrm{2}\right)\left({L}+\mathrm{1}\right)=\mathrm{0} \\ $$$${L}=\mathrm{2}\:\:\mid\:\:{L}=−\mathrm{1} \\ $$$$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+...}}}}\:\:\:{may}\:{not}\:{be}\:{negative} \\ $$$${and}\:\:{there}\:{is}\:{also}\:{chance}\:{of}\:{extraneous} \\ $$$${roots},\:{so}\:{I}\:{think}\:−\mathrm{1}\:{is}\:{extraneous}\:{root}. \\ $$$${So}\:{L}=\mathrm{2} \\ $$

Commented by FilupSmith last updated on 09/Feb/16

sorry i meant  a_1 =(√2)

$${sorry}\:{i}\:{meant} \\ $$$${a}_{\mathrm{1}} =\sqrt{\mathrm{2}} \\ $$

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