Compute $\sum\limits_{n=1}^{\infty} \frac1{1+x_n}$ if $x_1>0$ and $ x_{n+1} = x_n ^2 + x_n $

Question

Suppose $ x_1\in \Re $ , and let $ x_{n+1} = x_n ^2 + x_n $ for $ n\geq 1 $.

Assuming $ x_1 > 0 $, find

$ \sum_{n=1}^{\infty} 1/(1+x_n) $.

What can you say about the cases where $ x_1 < 0 $?

Notes: Just some help with evaluating the initial recurrence relation would be much appreciated.

By setting $ T_n = x_n + 1/2 $ I obtained the possibly simpler recurrence $ T_{n+1} = T_n^2 + 1/4 $.

Answer 1

Hint: \begin{align} \frac{1}{x_{n+1}}-\frac{1}{x_n}&=\frac{x_n- x_{n+1}}{x_n x_{n+1}}\\ &=\frac{x_n - x_n - x_n^2}{x_n x_{n+1}}\\ &=-\frac{x_n}{x_{n+1}}\\ &=-\frac{x_n}{x_n+x_n^2}\\ &=-\frac{1}{1+x_n} \end{align}