S-1-2-3-4-S-

Question Number 4285 by Filup last updated on 07/Jan/16

$${S}=\mathrm{1}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}+\sqrt{\mathrm{4}+…}}} \\ $$$${S}=??? \\ $$

Commented by RasheedSindhi last updated on 07/Jan/16

S=(√(2+(√(3+(√(4+...)))))) S(n)=(√((n+1)+S(n+1))) S(1)=(√(2+S(2))) S(2)=(√(3+S(3)) S(3)=(√(4+S(4)))

$$\mathrm{S}=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}+\sqrt{\mathrm{4}+…}}} \\ $$$${S}\left({n}\right)=\sqrt{\left({n}+\mathrm{1}\right)+{S}\left({n}+\mathrm{1}\right)} \\ $$$${S}\left(\mathrm{1}\right)=\sqrt{\mathrm{2}+{S}\left(\mathrm{2}\right)} \\ $$$${S}\left(\mathrm{2}\right)=\sqrt{\mathrm{3}+{S}\left(\mathrm{3}\right.} \\ $$$${S}\left(\mathrm{3}\right)=\sqrt{\mathrm{4}+{S}\left(\mathrm{4}\right)} \\ $$

Commented by Rasheed Soomro last updated on 10/Jan/16

A Useless Try S(n)=(√((n+1)+S(n+1))) S(n−1)=(√(n+S(n))) ( S(n) )^2 =(n+1)+S(n+1) ( S(n−1) )^2 =n+S(n) ( S(n) )^2 −( S(n−1) )^2 =1+S(n+1)−S(n) ( S(n) )^2 +S(n)−( S(n−1) )^2 −1−S(n+1)=0...............∗ By Quadratic formula S(n)=((−1±(√((1)^2 −4(1){−( S(n−1) )^2 −1−S(n+1)})))/(2(1))) S(n)=((−1±(√(1+4( S(n−1) )^2 +4+4S(n+1))))/(2(1))) 2S(n)=−1±(√(5+4( S(n−1) )^2 +4S(n+1))) 2S(n)+1=±(√(5+4( S(n−1) )^2 +4S(n+1))) (2S(n)+1)^2 =5+4( S(n−1) )^2 +4S(n+1) ...... .... S(n)=(√((n+1)+S(n+1))) S(n−1)=(√(n+S(n))) ( S(n−1) )^2 =n+S(n) S(n)=( S(n−1) )^2 −n

$$\mathbb{A}\:\mathbb{U}\mathrm{seless}\:\mathbb{T}\mathrm{ry} \\ $$$${S}\left({n}\right)=\sqrt{\left({n}+\mathrm{1}\right)+{S}\left({n}+\mathrm{1}\right)} \\ $$$${S}\left({n}−\mathrm{1}\right)=\sqrt{{n}+{S}\left({n}\right)} \\ $$$$\left(\:{S}\left({n}\right)\:\right)^{\mathrm{2}} =\left({n}+\mathrm{1}\right)+{S}\left({n}+\mathrm{1}\right) \\ $$$$\left(\:{S}\left({n}−\mathrm{1}\right)\:\right)^{\mathrm{2}} ={n}+{S}\left({n}\right) \\ $$$$\left(\:{S}\left({n}\right)\:\right)^{\mathrm{2}} −\left(\:{S}\left({n}−\mathrm{1}\right)\:\right)^{\mathrm{2}} =\mathrm{1}+{S}\left({n}+\mathrm{1}\right)−{S}\left({n}\right) \\ $$$$\left(\:{S}\left({n}\right)\:\right)^{\mathrm{2}} +{S}\left({n}\right)−\left(\:{S}\left({n}−\mathrm{1}\right)\:\right)^{\mathrm{2}} −\mathrm{1}−{S}\left({n}+\mathrm{1}\right)=\mathrm{0}……………\ast \\ $$$${By}\:{Quadratic}\:{formula} \\ $$$${S}\left({n}\right)=\frac{−\mathrm{1}\pm\sqrt{\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left\{−\left(\:{S}\left({n}−\mathrm{1}\right)\:\right)^{\mathrm{2}} −\mathrm{1}−{S}\left({n}+\mathrm{1}\right)\right\}}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$${S}\left({n}\right)=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}\left(\:{S}\left({n}−\mathrm{1}\right)\:\right)^{\mathrm{2}} +\mathrm{4}+\mathrm{4}{S}\left({n}+\mathrm{1}\right)}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$$\mathrm{2}{S}\left({n}\right)=−\mathrm{1}\pm\sqrt{\mathrm{5}+\mathrm{4}\left(\:{S}\left({n}−\mathrm{1}\right)\:\right)^{\mathrm{2}} +\mathrm{4}{S}\left({n}+\mathrm{1}\right)} \\ $$$$\mathrm{2}{S}\left({n}\right)+\mathrm{1}=\pm\sqrt{\mathrm{5}+\mathrm{4}\left(\:{S}\left({n}−\mathrm{1}\right)\:\right)^{\mathrm{2}} +\mathrm{4}{S}\left({n}+\mathrm{1}\right)} \\ $$$$\left(\mathrm{2}{S}\left({n}\right)+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{5}+\mathrm{4}\left(\:{S}\left({n}−\mathrm{1}\right)\:\right)^{\mathrm{2}} +\mathrm{4}{S}\left({n}+\mathrm{1}\right) \\ $$$$…… \\ $$$$…. \\ $$$${S}\left({n}\right)=\sqrt{\left({n}+\mathrm{1}\right)+{S}\left({n}+\mathrm{1}\right)} \\ $$$${S}\left({n}−\mathrm{1}\right)=\sqrt{{n}+{S}\left({n}\right)} \\ $$$$\left(\:{S}\left({n}−\mathrm{1}\right)\:\right)^{\mathrm{2}} ={n}+{S}\left({n}\right) \\ $$$${S}\left({n}\right)=\left(\:{S}\left({n}−\mathrm{1}\right)\:\right)^{\mathrm{2}} −{n} \\ $$

Commented by Rasheed Soomro last updated on 13/Jan/16

Second line should be S_n =S_(n−1) ^2 −n instead of S_n =S_(n+1) ^2 −n

$${Second}\:{line}\:{should}\:{be} \\ $$$${S}_{{n}} ={S}_{{n}−\mathrm{1}} ^{\mathrm{2}} −{n} \\ $$$${instead}\:{of}\: \\ $$$${S}_{{n}} ={S}_{{n}+\mathrm{1}} ^{\mathrm{2}} −{n} \\ $$

Commented by Filup last updated on 13/Jan/16

S_n =(√((n+1)+S_(n+1) )) S_n =S_(n+1) ^2 −n ∴(n+1)+S_(n+1) =(S_(n+1) ^( 2) −n) (n+1)+S_(n+1) =S_(n+1) ^( 4) −2S_(n+1) ^( 2) n+n^2 S_(n+1) ^( 4) −2nS_(n+1) ^( 2) −S_(n+1) +n^2 −n−1=0

$${S}_{{n}} =\sqrt{\left({n}+\mathrm{1}\right)+{S}_{{n}+\mathrm{1}} } \\ $$$${S}_{{n}} ={S}_{{n}+\mathrm{1}} ^{\mathrm{2}} −{n} \\ $$$$\therefore\left({n}+\mathrm{1}\right)+{S}_{{n}+\mathrm{1}} =\left({S}_{{n}+\mathrm{1}} ^{\:\mathrm{2}} −{n}\right) \\ $$$$\left({n}+\mathrm{1}\right)+{S}_{{n}+\mathrm{1}} ={S}_{{n}+\mathrm{1}} ^{\:\mathrm{4}} −\mathrm{2}{S}_{{n}+\mathrm{1}} ^{\:\mathrm{2}} {n}+{n}^{\mathrm{2}} \\ $$$${S}_{{n}+\mathrm{1}} ^{\:\mathrm{4}} −\mathrm{2}{nS}_{{n}+\mathrm{1}} ^{\:\mathrm{2}} −{S}_{{n}+\mathrm{1}} +{n}^{\mathrm{2}} −{n}−\mathrm{1}=\mathrm{0} \\ $$

Commented by Filup last updated on 11/Jan/16

S(n)=[S(n−1)]^2 −n or S_n =S_(n−1) ^( 2) −n What is the limit as n→∞?

$${S}\left({n}\right)=\left[{S}\left({n}−\mathrm{1}\right)\right]^{\mathrm{2}} −{n} \\ $$$${or} \\ $$$${S}_{{n}} ={S}_{{n}−\mathrm{1}} ^{\:\mathrm{2}} −{n} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{as}\:{n}\rightarrow\infty? \\ $$

Commented by Rasheed Soomro last updated on 11/Jan/16

I thought for that but don′t know how to determine it.

$$\mathrm{I}\:\mathrm{thought}\:\mathrm{for}\:\mathrm{that}\:\mathrm{but}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{how} \\ $$$$\mathrm{to}\:\mathrm{determine}\:\mathrm{it}. \\ $$

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