(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


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

${(a_{n})}_{n \in N} s u c h t h a t a_{1} = \frac{1}{2}$ $a n d a_{n + 1} = \frac{a_{n}^{2}}{a_{n}^{2} - a_{n} + 1}$ $P r o v e t h a t a_{1} + a_{2} + a_{3} + . . . + a_{n} < 1$ $n e e d h e l p e r$
Commented by sou1618 last updated on 15/Sep/16
$a_1 =(1/2) a_2 =(((1/2)^2 )/((1/2)^2 −(1/2)+1))=(((1/4))/((3/4)))=(1/3) .... ∗prove a_n <1 (mathematical induction) when n=1 a_1 =(1/2)<1 when n=k assume a_k <1 when n=k+1 a_(k+1) =(a_k ^2 /(a_k ^2 −a_k +1))=1+((a_k −1)/(a_k ^2 −a_k +1)) ((a_k −1)/(a_k ^2 −a_k +1))<0 (∵ { ((a_k −1<0)),((a_k ^2 −a_k +1>0)) :}) ⇒a_(k+1) <1 so a_n <1 (n∈N) −−−−−−−−−−−−−−−− a_(n+1) =(a_n ^2 /(a_n ^2 −a_n +1))=1+((a_n −1)/(a_n ^2 −a_n +1)) a_(n+1) −1=((a_n −1)/(a_n ^2 −a_n +1)) (1/(a_(n+1) −1))=((a_n ^2 −a_n +1)/(a_n −1)) (a_n ≠1) (1/(a_(n+1) −1))=a_n +(1/(a_n −1)) (1/(a_(n+1) −1))−(1/(a_n −1))=a_n S_n =a_1 +a_2 +...+a_(n−1) +a_n S_n =((1/(a_2 −1))−(1/(a_1 −1)))+((1/(a_3 −1))−(1/(a_2 −1)))+...+((1/(a_n −1))−(1/(a_(n−1) −1)))+((1/(a_(n+1) −1))−(1/(a_n −1))) S_n =−(1/(a_1 −1))+(1/(a_(n+1) −1)) S_n =2+(1/(a_(n+1) −1)) S_n =2−(1/(1−a_(n+1) )) (1/(1−a_(n+1) ))>(1/1) (∵ a_(n+1) <1) S_n <1$
$a_{1} = \frac{1}{2}$ $a_{2} = \frac{{(1 / 2)}^{2}}{{(1 / 2)}^{2} - (1 / 2) + 1} = \frac{(1 / 4)}{(3 / 4)} = \frac{1}{3}$ $. . . .$ $* p r o v e a_{n} < 1 (m a t h e m a t i c a l i n d u c t i o n)$ $w h e n n = 1$ $a_{1} = \frac{1}{2} < 1$ $w h e n n = k$ $a s s u m e a_{k} < 1$ $w h e n n = k + 1$ $a_{k + 1} = \frac{a_{k}^{2}}{a_{k}^{2} - a_{k} + 1} = 1 + \frac{a_{k} - 1}{a_{k}^{2} - a_{k} + 1}$ $\frac{a_{k} - 1}{a_{k}^{2} - a_{k} + 1} < 0 (∵ {\begin{cases} a_{k} - 1 < 0 \\ a_{k}^{2} - a_{k} + 1 > 0 \end{cases})$ $\Rightarrow a_{k + 1} < 1$ $s o$ $a_{n} < 1 (n \in N)$ $- - - - - - - - - - - - - - - -$ $a_{n + 1} = \frac{a_{n}^{2}}{a_{n}^{2} - a_{n} + 1} = 1 + \frac{a_{n} - 1}{a_{n}^{2} - a_{n} + 1}$ $a_{n + 1} - 1 = \frac{a_{n} - 1}{a_{n}^{2} - a_{n} + 1}$ $\frac{1}{a_{n + 1} - 1} = \frac{a_{n}^{2} - a_{n} + 1}{a_{n} - 1} (a_{n} \neq 1)$ $\frac{1}{a_{n + 1} - 1} = a_{n} + \frac{1}{a_{n} - 1}$ $\frac{1}{a_{n + 1} - 1} - \frac{1}{a_{n} - 1} = a_{n}$ $S_{n} = a_{1} + a_{2} + . . . + a_{n - 1} + a_{n}$ $S_{n} = (\frac{1}{a_{2} - 1} - \frac{1}{a_{1} - 1}) + (\frac{1}{a_{3} - 1} - \frac{1}{a_{2} - 1}) + . . . + (\frac{1}{a_{n} - 1} - \frac{1}{a_{n - 1} - 1}) + (\frac{1}{a_{n + 1} - 1} - \frac{1}{a_{n} - 1})$ $S_{n} = - \frac{1}{a_{1} - 1} + \frac{1}{a_{n + 1} - 1}$ $S_{n} = 2 + \frac{1}{a_{n + 1} - 1}$ $S_{n} = 2 - \frac{1}{1 - a_{n + 1}}$ $\frac{1}{1 - a_{n + 1}} > \frac{1}{1} (∵ a_{n + 1} < 1)$ $S_{n} < 1$
Commented by Chantria last updated on 16/Sep/16

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