Finding the sum of an inverse series define by recursion

Question

We define a series by a recursion

x_1=a; x_(n+1)=(x_n)^2-x_n+1

Find an expression for the sum 1/x_n (n goes from 1 to infinity)

See the pic

Answer 1

Notice that $$1/(x_n-1) = 1/x_n + 1/(x_{n+1}-1)$$ for all $n$. Thus $$ 1/(a-1) = 1/(x_1-1) = \sum_{n=1}^\infty 1/x_n.$$ In other words, this is a Telescoping series.

Answer 2

In the recurrence formula, you get $$x_{n+1}-1=x_n(x_n-1)$$ Then $$\frac{1}{x_{n+1}-1}=\frac{1}{x_n(x_n-1)}=\frac{1}{x_n-1}-\frac{1}{x_n}$$ Therefore $$\begin{align}\sum_{n=1}^{\infty}\frac{1}{x_n}=&\sum_{n=1}^{\infty}\left(\frac{1}{x_n-1}-\frac{1}{x_{n+1}-1}\right)\\ =& \lim_{N\to \infty}\sum_{n=1}^N\left(\frac{1}{x_n-1}-\frac{1}{x_{n+1}-1}\right)\\ =&\lim_{N\to \infty}\left(\frac{1}{x_1-1}+\sum_{n=2}^N\frac{1}{x_n-1}-\sum_{n=1}^N\frac{1}{x_{n+1}-1}\right)\\ =&\lim_{N\to \infty}\left(\frac{1}{x_1-1}-\frac{1}{x_{N+1}-1}\right)\\=&\,\frac{1}{a-1}\end{align}$$ whenever $x_N\to\infty$ as $N\to\infty$. $(a\neq 1)$