How to find the sum of the number series written recursively? I have the following number series, written in next way: 
$x_1 = 2$ 
$x_{n+1} = x_n^2 - x_n + 1$, $n \geq 1$
and it needs to find sum of: 
$$\sum_{n=1}^{\infty} \frac{1}{x_n}$$
At the moment, I haven't had much of progress, namely: 
I know, that in the end of solving this task I'll have next type of situation:
 $(x_1+x_2) - (x_2 +x_3) - (x_3 + x_4) ... (x_{n-1} + x_n) $ 
so elements mutually cancel out inside the sum and we get desired result
But every action requires mathematical proof unless it is obviously true
I've had also next observation: 
$x_n$ is growing, thus $\frac{1}{x_n}$ decreases, power of decreasing $\gt 1$ thus the number series is convergent, but yet again, I don't have strictly math proof of this fact, but it requires
I'll be very grateful for any help

Duplicate of this question, found on approach0.xyz.
But this question doesn't have any answer, so I'll just share the important hint given in the comments by @Kelenner:
$$
\frac{1}{x_n -1} - \frac{1}{x_{n + 1} -1} 
= \frac{1}{x_n}
$$
As you detailed in the comments, you have used the hint to show the sum is equal to $\frac{1}{x_1 - 1}$.
For the sake of completeness, I'll write it down here.
First, we have
\begin{align}
\frac{1}{x_n -1} - \frac{1}{x_{n + 1} -1} 
& = \frac{1}{x_n -1} - \frac{1}{x_n^2 - x_n + 1 -1} \\
& = \frac{x_n}{x_n^2 -x_n} - \frac{1}{x_n^2 - x_n} \\
& = \frac{x_n - 1}{x_n^2 -x_n}
= \frac{1}{x_n}.
\end{align}
Therefore, we obtain by a telescoping argument
\begin{align}
\sum_{n = 1}^{\infty} \frac{1}{x_{n}}
& = \sum_{n = 1}^{\infty} \frac{1}{x_n -1} - \frac{1}{x_{n + 1} -1} 
= \lim_{m \to \infty} \left(\sum_{n = 1}^{m} \frac{1}{x_n - 1} - \frac{1}{x_{n + 1} - 1}\right) \\
& = \lim_{m \to \infty}\left( \frac{1}{x_1 - 1} - \frac{1}{x_{m + 1} - 1}\right)
= \frac{1}{x_1 - 1} + \lim_{n \to \infty} \frac{-1}{x_{n + 1} -1} 
= \frac{1}{x_1 - 1}
\end{align}
As we have $x_n \to \infty$.
This can be seen by showing $x_n \ge n$ by induction.
For $n = 1$ its clear, so we begin at $n = 2$.
Base step: $x_2 = 3 \ge 2$.
Induction hypothesis: $x_n \ge n$ holds for some $n \in \mathbb{N}_{\ge 2}$.
Induction step: $n \to n + 1$.
Since $x_n \ge 2$ for all $n$ we have
$$
x_{n + 1}
= x_n(x_n - 1) + 1
\overset{\text{IH}}{\ge} (n)(n - 1) + 1
= n^2 - n + 1 \ge n + 1.
$$
The last inequality rearranges to $n^2 \ge 2n \iff n \ge 2$. So the statement is true for $n \ge 2$ but we know its also true for $n = 1$, so we are done.
