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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 $x_1=\frac {1}{100}, x_n={x_{n1}}^2+2x_{n1}, n\ge2$ 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 $(1x_n)=(1x_{n1})^2$ 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 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra  Let $y_n=1x_n$. Then $y_n=y_{n1}^2$ and the formula is easily found to be $$y_n\ =\ y_1^{2^{n1}}$$ i.e. $1x_n \, = \, (1x_1)^{2^{n1}}$ $\implies$ $x_n \, = \, 1(1x_1)^{2^{n1}} \, = \, 1  \left(\dfrac{99}{100}\right)^{2^{n1}}$. Try plugging in this formula and see. 

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