Show the series is convergent and find the limit Sequence of real numbers $a_n$ defined recursively with $a_1=1/2$ and $a_{n+1}=\frac{a_n^2}{a_n^2-a_n+1}$  for all $n \geq 1$.
Show that $\sum_{n=1}^\infty a_n$ is convergent and find its limit.
I have tried to convert the recursive form to explicit form but it's too difficult.

Define the following sequence $b_n = \frac{1}{a_n}$. Then
$$b_{n+1} =
\frac{a_n^2-a_n+1}{a_n^2} = 1 - \frac{1}{a_n}+ \frac{1}{a_n^2} = b_n^2-b_n+1
$$
so the sequence $b_n$ is an increasing sequence of natural numbers whose first terms are
$$2, \ 3, \ 7, \ 43, \ 1807, \ \dots$$
Partial sums have a curious pattern: if you denote $s_n = a_1 + \cdots + a_n$ you have


*

*$s_1 = \frac{1}{2} = 1 - \frac{1}{b_1} = 1- \frac{1}{b_2-1}$

*$s_2 = \frac{5}{6} = 1- \frac{1}{b_1b_2} = 1- \frac{1}{b_3-1}$

*$s_3 = \frac{41}{42} = 1- \frac{1}{b_1b_2b_3} = 1- \frac{1}{b_4-1}$

*$s_4 = \frac{1805}{1806} = 1- \frac{1}{b_1b_2b_3b_4} = 1- \frac{1}{b_5-1}$
So it seems that the sum of the series is $1$ and the partial sums are $s_n=1- a_1 \cdots a_n$
EDIT: Wikipedia says that this is called Sylvester Sequence. See http://en.wikipedia.org/wiki/Sylvester%27s_sequence
