Divergent series

[rtz2k4]

[math]f:\mathbb{R}\backslash \left\{\sqrt{3}\right\}\rightarrow \mathbb{R},\:f\left(x\right)=\frac{x\sqrt{3}+1}{\sqrt{3}-x}[/math]
[math]\left(a_n\right),n\ge \:1,a_1=2,a_{n+1}=f\left(a_n\right),n\epsilon \mathbb{N^{*}}[/math]
How do I prove [imath]a_{n}[/imath] is not a convergent series? I need to calculate his limit and it must be infinite or minus infinite or the limit doesn't exist. How do I calculate the limit lim(n->inf)(a_1+a_2+a_3+...+(a_(n+1))? should i transform it to lim(n->inf)(2+f(a_1)+f(a_2)+...+(f(a_n)) and after that I replace f(a_n) lim(n->inf)((2*sqrt(3)+1)/((sqrt(3)-2)+((a_1)*sqrt(3)+1)/(sqrt(3)-(a_1))+...+((a_n)*sqrt(3)+1)/(sqrt(3)-(a_n)).

 

[JeffM]

I have not worked this out, but it may be helpful here to rationalize the denominator.

[math] f(x) = \dfrac{x \sqrt{3} + 1}{\sqrt{3} - x} = \dfrac{x \sqrt{3} + 1}{\sqrt{3} - x} * \dfrac{x + \sqrt{3}}{\sqrt{3} + x} = - \dfrac{x^2 \sqrt{3} + 4x + \sqrt{3}}{x^2 - 3}. [/math]

 

[Steven G]

Are all the terms positive or negative after some point? If yes, then one way of showing that a series is divergent is to show that the end terms are not approaching 0. Did you try that? That's what I would try first.

 

[blamocur]

Try figuring out [imath]a_n[/imath] for [imath]1 < n \leq 7[/imath]