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 May 12th, 2014, 06:01 PM #1 Newbie   Joined: May 2014 From: korea Posts: 1 Thanks: 0 inequality about sum Sequence $(x_n)$ is difined $x_1=\frac {1}{100}, x_n=-{x_{n-1}}^2+2x_{n-1}, n\ge2$ Prove that $$\sum_{n=1}^\infty [(x_{n+1}-x_n)^2+(x_{n+1}-x_n)(x_{n+2}-x_{n+1})]\lt \frac {1}{3}$$ I found relation $(1-x_n)=(1-x_{n-1})^2$ I don't know what to do next. There is a real number which is less than $\frac {1}{3}$? I need your help. May 13th, 2014, 05:37 AM   #2
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Quote:
 Originally Posted by mmmathhh I found relation $(1-x_n)=(1-x_{n-1})^2$
Let $y_n=1-x_n$. Then $y_n=y_{n-1}^2$ and the formula is easily found to be
$$y_n\ =\ y_1^{2^{n-1}}$$
i.e. $1-x_n \, = \, (1-x_1)^{2^{n-1}}$ $\implies$ $x_n \, = \, 1-(1-x_1)^{2^{n-1}} \, = \, 1 - \left(\dfrac{99}{100}\right)^{2^{n-1}}$. Try plugging in this formula and see. Tags inequality, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Ionika Algebra 2 February 2nd, 2014 06:53 PM gaussrelatz Algebra 15 October 27th, 2012 02:58 PM Pavhard Number Theory 4 October 18th, 2010 07:00 AM tinynerdi Calculus 4 September 27th, 2010 04:09 PM elim Elementary Math 0 April 15th, 2010 04:45 PM

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