Assume-that-the-sequence-terms-tend-to-the-constant-value-u-so-that-as-n-u-n-1-u-and-u-n-u-i-show-that-u-2-u-1-0-ii-show-that-1-1-1-1-1-1-1-1-1-5-2-

Question Number 175211 by MathsFan last updated on 23/Aug/22

$$\boldsymbol{\mathrm{Assume}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sequence}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{tend}} \\ $$$$\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{constant}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{{u}},\:\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{as}} \\ $$$$\boldsymbol{\mathrm{n}}\rightarrow\infty,\:\boldsymbol{{u}}_{\boldsymbol{{n}}−\mathrm{1}} \rightarrow\boldsymbol{{u}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{{u}}_{\boldsymbol{{n}}} \rightarrow\boldsymbol{{u}}. \\ $$$$\:\left(\boldsymbol{\mathrm{i}}\right)\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\:\boldsymbol{{u}}^{\mathrm{2}} +\boldsymbol{{u}}−\mathrm{1}=\mathrm{0} \\ $$$$\:\left(\boldsymbol{\mathrm{ii}}\right)\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+…..}}}}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Answered by Rasheed.Sindhi last updated on 23/Aug/22

$$\left(\boldsymbol{\mathrm{ii}}\right)\:\boldsymbol{\mathrm{u}}=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+…..}}}};\:\boldsymbol{\mathrm{u}}>\mathrm{0} \\ $$$$\:\:\:\boldsymbol{\mathrm{u}}=\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{\mathrm{u}}} \\ $$$$\boldsymbol{\mathrm{u}}^{\mathrm{2}} +\boldsymbol{\mathrm{u}}−\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{u}}=\frac{−\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{1}\right)}}{\mathrm{2}}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left[\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}<\mathrm{0}\right] \\ $$

Commented by MathsFan last updated on 24/Aug/22

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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