Sequence and series with specific nth term Let $a(1) = 2$ , $a(n+1) = a(n)^2-a(n) + 1$  for $n\geq 1$, Find $$\sum_{n=1}^{\infty} \frac{1}{a(n)}$$ 

Probably not an answer.
As said in comments, computing the very first values of $a_n$, you get
$$\{2,3,7,43,1807,3263443\}$$ and the corresponding sums of their reciprocals are then
$$\left\{\frac{1}{2},\frac{5}{6},\frac{41}{42},\frac{1805}{1806},\frac{3263441}{32
   63442},\frac{10650056950805}{10650056950806}\right\}$$ where you could notice some very interesting patterns.
What it seems is that 
$$\sum_{n=1}^{p} \frac{1}{a_n}=\frac{a_{p+1}-2}{a_{p+1}-1}=1-\frac{1}{a_{p+1}-1}$$
It looks like the greedy Egyptian representation of $1$ !
