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
 A: 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.
A: To prove the limit is $1$, prove two results by induction, firstly that $x_{n+1}=1+\prod_{j=1}^nx_j$, and secondly that $\sum_{j=1}^n\frac{1}{x_j}=1-\frac{1}{\prod_{j=1}^nx_j}=1-\frac{1}{x_{n+1}-1}$. The first proof's inductive step is$$x_{k+1}=1+\prod_{j=1}^kx_j\implies x_{k+2}=1+(x_{k+1}-1)x_{k+1}=1+\prod_{j=1}^{k+1}x_j,$$while the second's is$$\sum_{j=1}^k\frac{1}{x_j}=1-\frac{1}{x_{k+1}-1}\implies\sum_{j=1}^{k+1}\frac{1}{x_j}=1-\frac{1}{x_{k+1}(x_{k+1}-1)}=1-\frac{1}{x_{k+2}-1}.$$
