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 $$\sum_{n=1}^{\infty} \frac{1}{a(n)}$$ 
 A: We first prove the following by induction:
$$
\frac{1}{a(n + 1)-1} + \sum_{k = 1}^n \frac{1}{a(k)} = 1.
$$
For $n = 1$, there stands
$$
\frac{1}{4 - 2 + 1 - 1} + \frac{1}{2} = 1,
$$
which holds. Suppose it holds for $n$. Then,
\begin{align*}
\frac{1}{a(n + 2) - 1} + \sum_{k = 1}^{n + 1} \frac{1}{a(k)} &= \frac{1}{a(n + 2) - 1} + \frac{1}{a(n + 1)} - \frac{1}{a(n + 1) - 1} + \frac{1}{a(n + 1) - 1} + \sum_{k = 1}^n \frac{1}{a(k)} \\
&= \frac{1}{a(n + 1)^2 - a(n + 1) + 1 - 1} + \frac{a(n + 1) - 1 - a(n + 1)}{a(n + 1)(a(n + 1) - 1)} + 1 \\
&= 1.
\end{align*}
Consequently,
$$
\sum_{k = 1}^\infty \frac{1}{a(k)} = \lim_{n \to \infty} \sum_{k = 1}^n \frac{1}{a(k)} = \lim_{n \to \infty} \left(1 - \frac{1}{a(n + 1) - 1}\right) = 1,
$$
since $a(n) \to \infty$ for $n \to \infty$.

EDIT: A little bit of explanation as to how I came up with the first step. I encountered this problem a while ago in a pdf on problem solving, but there was an extra part: it asked to show that $a(n)$ and $a(m)$ are coprime if $n \ne m$. To show that, it naturally came up to look at $a(n) - 1$. That was the first part of the problem, the second part was finding the series of the reciprocals. While trying to solve it, I calculated $\frac{1}{a(n)} + \frac{1}{a(n + 1)}$. When that did not give anything useful, I tried doing something with $\frac{1}{a(n + 1) - 1}$, as this kind of term was used in the first part.
Hopefully this gives a bit of an idea how I came up with my solution.
A: (not an answer)
I calculated with Python:
def f(x): return x**2 -x +1

def list_sum(a):
l = []
term = a
for k in range(20):
    l.append(1.0/term)
    term = min(f(term),50000000000)
    print "term=", term
return sum(l)

print list_sum(2)

and the answer seems to be quite close to $1$.
A: 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 $1$ !
