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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} + \dots + a_{n} < 1

n e e d h e l p e r

Commented by sou1618 last updated on 15/Sep/16

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}

\dots .

* 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} + \dots + 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}) + \dots + (\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

n i c e s i r . t h a n k s