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May 12th, 2014, 06:01 PM   #1
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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.
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May 13th, 2014, 05:37 AM   #2
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Originally Posted by mmmathhh View Post
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.
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