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Question Number 7788 by Chantria last updated on 15/Sep/16

(a_n )_(n∈N) such that a_1 =(1/2) and a_(n+1) =(a_n ^2 /(a_n ^2 −a_n +1)) Prove that a_1 +a_2 +a_3 +...+a_n < 1 need helper

$$\left({a}_{{n}} \right)_{{n}\in{N}} \:{such}\:{that}\:{a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$ $${and}\:{a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} ^{\mathrm{2}} }{{a}_{{n}} ^{\mathrm{2}} −{a}_{{n}} +\mathrm{1}} \\ $$ $${Prove}\:{that}\:{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +...+{a}_{{n}} \:<\:\mathrm{1} \\ $$ $${need}\:{helper} \\ $$ $$ \\ $$

Commented bysou1618 last updated on 15/Sep/16

a_1 =(1/2) a_2 =(((1/2)^2 )/((1/2)^2 −(1/2)+1))=(((1/4))/((3/4)))=(1/3) .... ∗prove a_n <1 (mathematical induction) when n=1 a_1 =(1/2)<1 when n=k assume a_k <1 when n=k+1 a_(k+1) =(a_k ^2 /(a_k ^2 −a_k +1))=1+((a_k −1)/(a_k ^2 −a_k +1)) ((a_k −1)/(a_k ^2 −a_k +1))<0 (∵ { ((a_k −1<0)),((a_k ^2 −a_k +1>0)) :}) ⇒a_(k+1) <1 so a_n <1 (n∈N) −−−−−−−−−−−−−−−− a_(n+1) =(a_n ^2 /(a_n ^2 −a_n +1))=1+((a_n −1)/(a_n ^2 −a_n +1)) a_(n+1) −1=((a_n −1)/(a_n ^2 −a_n +1)) (1/(a_(n+1) −1))=((a_n ^2 −a_n +1)/(a_n −1)) (a_n ≠1) (1/(a_(n+1) −1))=a_n +(1/(a_n −1)) (1/(a_(n+1) −1))−(1/(a_n −1))=a_n S_n =a_1 +a_2 +...+a_(n−1) +a_n S_n =((1/(a_2 −1))−(1/(a_1 −1)))+((1/(a_3 −1))−(1/(a_2 −1)))+...+((1/(a_n −1))−(1/(a_(n−1) −1)))+((1/(a_(n+1) −1))−(1/(a_n −1))) S_n =−(1/(a_1 −1))+(1/(a_(n+1) −1)) S_n =2+(1/(a_(n+1) −1)) S_n =2−(1/(1−a_(n+1) )) (1/(1−a_(n+1) ))>(1/1) (∵ a_(n+1) <1) S_n <1

$${a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $${a}_{\mathrm{2}} =\frac{\left(\mathrm{1}/\mathrm{2}\right)^{\mathrm{2}} }{\left(\mathrm{1}/\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{1}/\mathrm{2}\right)+\mathrm{1}}=\frac{\left(\mathrm{1}/\mathrm{4}\right)}{\left(\mathrm{3}/\mathrm{4}\right)}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$ $$.... \\ $$ $$\ast{prove}\:\:{a}_{{n}} <\mathrm{1}\:\left({mathematical}\:{induction}\right) \\ $$ $${when}\:{n}=\mathrm{1} \\ $$ $$\:\:\:\:\:{a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}<\mathrm{1} \\ $$ $${when}\:{n}={k} \\ $$ $$\:\:\:\:\:{assume}\:{a}_{{k}} <\mathrm{1} \\ $$ $${when}\:{n}={k}+\mathrm{1} \\ $$ $$\:\:\:\:{a}_{{k}+\mathrm{1}} =\frac{{a}_{{k}} ^{\mathrm{2}} }{{a}_{{k}} ^{\mathrm{2}} −{a}_{{k}} +\mathrm{1}}=\mathrm{1}+\frac{{a}_{{k}} −\mathrm{1}}{{a}_{{k}} ^{\mathrm{2}} −{a}_{{k}} +\mathrm{1}} \\ $$ $$\:\:\:\:\frac{{a}_{{k}} −\mathrm{1}}{{a}_{{k}} ^{\mathrm{2}} −{a}_{{k}} +\mathrm{1}}<\mathrm{0}\:\left(\because\begin{cases}{{a}_{{k}} −\mathrm{1}<\mathrm{0}}\\{{a}_{{k}} ^{\mathrm{2}} −{a}_{{k}} +\mathrm{1}>\mathrm{0}}\end{cases}\right) \\ $$ $$\:\:\:\:\:\Rightarrow{a}_{{k}+\mathrm{1}} <\mathrm{1} \\ $$ $${so} \\ $$ $${a}_{{n}} <\mathrm{1}\:\left({n}\in\mathbb{N}\right) \\ $$ $$−−−−−−−−−−−−−−−− \\ $$ $${a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} ^{\mathrm{2}} }{{a}_{{n}} ^{\mathrm{2}} −{a}_{{n}} +\mathrm{1}}=\mathrm{1}+\frac{{a}_{{n}} −\mathrm{1}}{{a}_{{n}} ^{\mathrm{2}} −{a}_{{n}} +\mathrm{1}} \\ $$ $${a}_{{n}+\mathrm{1}} −\mathrm{1}=\frac{{a}_{{n}} −\mathrm{1}}{{a}_{{n}} ^{\mathrm{2}} −{a}_{{n}} +\mathrm{1}} \\ $$ $$\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} −\mathrm{1}}=\frac{{a}_{{n}} ^{\mathrm{2}} −{a}_{{n}} +\mathrm{1}}{{a}_{{n}} −\mathrm{1}}\:\:\:\left({a}_{{n}} \neq\mathrm{1}\right) \\ $$ $$\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} −\mathrm{1}}={a}_{{n}} +\frac{\mathrm{1}}{{a}_{{n}} −\mathrm{1}} \\ $$ $$\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} −\mathrm{1}}−\frac{\mathrm{1}}{{a}_{{n}} −\mathrm{1}}={a}_{{n}} \\ $$ $$ \\ $$ $${S}_{{n}} ={a}_{\mathrm{1}} +{a}_{\mathrm{2}} +...+{a}_{{n}−\mathrm{1}} +{a}_{{n}} \\ $$ $${S}_{{n}} =\left(\frac{\mathrm{1}}{{a}_{\mathrm{2}} −\mathrm{1}}−\frac{\mathrm{1}}{{a}_{\mathrm{1}} −\mathrm{1}}\right)+\left(\frac{\mathrm{1}}{{a}_{\mathrm{3}} −\mathrm{1}}−\frac{\mathrm{1}}{{a}_{\mathrm{2}} −\mathrm{1}}\right)+...+\left(\frac{\mathrm{1}}{{a}_{{n}} −\mathrm{1}}−\frac{\mathrm{1}}{{a}_{{n}−\mathrm{1}} −\mathrm{1}}\right)+\left(\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} −\mathrm{1}}−\frac{\mathrm{1}}{{a}_{{n}} −\mathrm{1}}\right) \\ $$ $${S}_{{n}} =−\frac{\mathrm{1}}{{a}_{\mathrm{1}} −\mathrm{1}}+\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} −\mathrm{1}} \\ $$ $${S}_{{n}} =\mathrm{2}+\frac{\mathrm{1}}{{a}_{{n}+\mathrm{1}} −\mathrm{1}} \\ $$ $${S}_{{n}} =\mathrm{2}−\frac{\mathrm{1}}{\mathrm{1}−{a}_{{n}+\mathrm{1}} } \\ $$ $$\frac{\mathrm{1}}{\mathrm{1}−{a}_{{n}+\mathrm{1}} }>\frac{\mathrm{1}}{\mathrm{1}}\:\left(\because\:{a}_{{n}+\mathrm{1}} <\mathrm{1}\right) \\ $$ $${S}_{{n}} <\mathrm{1} \\ $$ $$ \\ $$

Commented byChantria last updated on 16/Sep/16

nice sir. thanks

$${nice}\:{sir}.\:{thanks} \\ $$

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