Find a sum of a convergent series Let $x_n$ be a sequence that is given by the following recursive formula:
$x_{n+1} = x_n^2 - x_n +1$, where $x_1=a \gt 1$. 
Find: $$\sum_{n=1}^{\infty} \frac{1}{x_n}$$
Not sure really how to approach this. I was perhaps thinking of using Tauber theorem, which would first require me to show the series in question is Abel summable. Would appreciate clues or just general insight. 
P.S. I am pretty sure this is not a telescoping series, so the obvious doesn't work.
 A: By playing around with the recursion, we see that after an insuspicious start, the sequence $x_n$ soon starts growing rapidly. For example, with $a=2$, we have:
$$ x_1=2, x_2= 3, x_3=7, x_4=43, x_5=1807,x_6=3263443,\ldots$$Substituting $x_n$ so that the recursion formula looks simple might be a good idea. For example,
$$x_{n+1}=\left(x_n-\frac12\right)^2+\frac 34 $$
might suggest to work with $y_n:=x_n-\frac12$ and
$$y_{n+1}=y_n^2+\frac14.$$
However, a different substitution is more helpful. The summand $\frac14$ is annoying, anything multiplicative would be more tractable. So we try to find out: For which $c$ is $x_{n+1}+c=x_n^2-x_n+1+c$ a nice multiple of $x_n+c$? It turns out (e.g., by polynomial division) that this is the case for $c:=-1$, i.e.,  with $z_n:=x_n-1$, we find
$$ \tag1z_{n+1}=z_nx_n=z_n(z_n+1).$$
So $z_n$ grows by a factor which itself grows (as we are about to see). More formally, we can readily use $(1)$ to show by induction that $z_n>0$ for all $n$. Using that and $(1)$, induction tells us $z_{n+1}>z_n$ for all $n$. Using that and $(1)$, induction tells us the (still very conservative) lower bound $$\tag2z_n\ge z_1\cdot a^{n-1}. $$
Taking reciprocals of $(1)$ leads to 
$$\frac1{z_{n+1}}=\frac1{z_n(z_n+1)}, $$
which - by good guessing or by explicitly looking for partial fractions - amounts to
$$\tag3\frac1{z_{n+1}}=\frac1{z_n}-\frac1{z_n+1}=\frac1{z_n}-\frac1{x_n}. $$
This way, we were lucky and could retrieve the desired $\frac1{x_n}$.
Summing $(3)$ from $n=1$ to $n=N$ gives
$$\sum_{n=1}^N\frac1{z_{n+1}}=\sum_{n=1}^N\frac1{z_n}-\sum_{n=1}^N\frac1{x_n}, $$
rearrange to
$$\sum_{n=1}^N\frac1{x_n} = \sum_{n=1}^N\frac1{z_n}-\sum_{n=1}^N\frac1{z_{n+1}} = \sum_{n=1}^N\frac1{z_n}-\sum_{n=2}^{N+1}\frac1{z_{n+1}}=\frac1{z_1}-\frac1{z_{N+2}}. $$
By $(2)$ and because $x_1>1$, this (rapidly) converges to $\frac1{z_1}=\frac1{x_1-1}$ as $N\to \infty$.
