Prove that the series$\sum a_n$, with $a_n$ defined recursively, converges to 1 Let $a_n$ , with $n\geq1$ be the sequence of real numbers defined by $a_1=\frac12$ and the following relation
$$a_{n+1}=\frac{a_n^2}{a_n^2-a_n+1}$$Prove that $\sum a_n$ converges to $1$.
Now, I have only been able to study the sequence defined by the recurrence relation. I know that it converges to zero by the monotone convergence theorem. I think I should prove that this is a geometric series like this $\sum_{n=0}^\infty \frac12\left(\frac12\right)^n$. What is the strategy I should follow in these cases? How would you approach this little problem?
Thank you for your help.

Let $S_n = \sum_{k=1}^n a_k$, then we observe the following pattern:
$$S_1 = \frac{1}{2}, a_2 = \frac{1}{3}$$
$$S_2 = \frac{5}{6}, a_3 = \frac{1}{7}$$
$$S_3 = \frac{41}{42}, a_4 = \frac{1}{43}$$
Actually, if $S_n = \dfrac{k-1}{k} = 1 -\dfrac{1}{k}$ and $a_{n+1} = \dfrac{1}{k+1}$ then $S_{n+1} = \dfrac{k^2 + k -1}{k^2 +k} = 1- \dfrac{1}{k^2 + k}$ and $a_{n+2} = \dfrac{1}{k(1 +k) + 1}$. 
ALl these suggest us to prove by induction $S_n = 1- \dfrac{1}{b_{n+1} - 1}$ and $a_{n+1} = \dfrac{1}{b_{n + 1}}$ with $b_{n+1} = b_n^2 - b_n + 1 $ and $b_1 =2$. It's easy to see $b_n \to +\infty$
This is obvious when $n = 1$. 
If this is true when $n = k$, i.e. if $S_k = 1- \dfrac{1}{b_{k+1} - 1}$ and $a_{k+1} = \dfrac{1}{b_{k + 1}}$ is true. then we can see $$a_{k+2}  = \dfrac{a_k^2}{a_k^2 - a_k + 1} = \dfrac{1}{b_{k+1}^2 - b_{k+1} + 1} = \dfrac{1}{b_{k+2} }$$ and $$S_{k+1} = S_k + a_{k+1} =1 - \dfrac{1}{b_{k+1}^2 - b_{k+1}}= 1 - \dfrac{1}{b_{k+2} - 1}$$
