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Question Number 153897 by mathdanisur last updated on 11/Sep/21

Denote  x_n   is the unique positive root  of the following equation:  x^n  + x^(n−1)  + ... x = n + 2  Prove that the sequence (x_n ) converges  to a positive real number. Find that  limit.

$$\mathrm{Denote}\:\:\mathrm{x}_{\boldsymbol{\mathrm{n}}} \:\:\mathrm{is}\:\mathrm{the}\:\mathrm{unique}\:\mathrm{positive}\:\mathrm{root} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{equation}: \\ $$$$\mathrm{x}^{\boldsymbol{\mathrm{n}}} \:+\:\mathrm{x}^{\boldsymbol{\mathrm{n}}−\mathrm{1}} \:+\:...\:\mathrm{x}\:=\:\mathrm{n}\:+\:\mathrm{2} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequence}\:\left(\mathrm{x}_{\boldsymbol{\mathrm{n}}} \right)\:\mathrm{converges} \\ $$$$\mathrm{to}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{number}.\:\mathrm{Find}\:\mathrm{that} \\ $$$$\mathrm{limit}. \\ $$

Commented by MJS_new last updated on 12/Sep/21

this is not a proof but...  f_n (x)=x^n +x^(n−1) +...+x  ⇒  f_n (1)=n  ⇒  f_n (1+k)>n∀k>0  ⇒  ∃k>0: f_n (1+k)=n+2  obviously  n→∞ ⇒ k→0

$$\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{proof}\:\mathrm{but}... \\ $$$${f}_{{n}} \left({x}\right)={x}^{{n}} +{x}^{{n}−\mathrm{1}} +...+{x} \\ $$$$\Rightarrow \\ $$$${f}_{{n}} \left(\mathrm{1}\right)={n} \\ $$$$\Rightarrow \\ $$$${f}_{{n}} \left(\mathrm{1}+{k}\right)>{n}\forall{k}>\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\exists{k}>\mathrm{0}:\:{f}_{{n}} \left(\mathrm{1}+{k}\right)={n}+\mathrm{2} \\ $$$$\mathrm{obviously} \\ $$$${n}\rightarrow\infty\:\Rightarrow\:{k}\rightarrow\mathrm{0} \\ $$

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