$y_0 \ge 2$, $y_n = y_{n-1}^2 -2$ $\Rightarrow$ $\frac{1}{y_0}+\frac{1}{y_0y_1}+\cdots = \frac{y_0 - \sqrt{y_0^2 - 4}}{2}$ $y_0 \ge 2$ and $y_n = y_{n-1}^2 - 2$. Let $S_n = \frac{1}{y_0} + \frac{1}{y_0 y_1}+\cdots + \frac{1}{y_0 y_1 \cdots y_{n}}$. Prove that
$$\lim_{n \rightarrow \infty} S_n = \frac{y_0 - \sqrt{y_0^2 - 4}}{2}$$
I don't have any idea to deal with it.
 A: $\text{let} P_n=y_0 y_1\text{...} y_n$, now I will show that $S_n^2-y_0 S_n+1=\frac{1}{P_n^2}$ by MI.
Base case $n=0$ is obvious, assume $S_k^2-y_0 S_k+1=\frac{1}{P_k^2}$, when $n=k+1$
\begin{align}
S_{k+1}^2-y_0 S_{k+1}+1 &= 
\left(S_k+\frac{1}{P_{k+1}}\right)^2-y_0 \left(S_k+\frac{1}{P_{k+1}}\right)+1 \\&=
S_k^2-y_0S_k+1+\frac{2 S_k}{P_{k+1}}-\frac{y_0}{P_{k+1}}+\frac{1}{P_{k+1}^2} \\ &=
\frac{1}{P_k^2}+\frac{2 S_k}{P_{k+1}}-\frac{y_0}{P_{k+1}}+\frac{1}{P_{k+1}^2}
\end{align}
Then we need $\frac{1}{P_k^2}+\frac{2 S_k}{P_{k+1}}-\frac{y_0}{P_{k+1}}=0$. Again, I will show it by MI.
Base case $n=0$ is obvious, assume $\frac{1}{P_m^2}+\frac{2 S_m}{P_{m+1}}-\frac{y_0}{P_{m+1}}=0$, when $k=m+1$
\begin{align}
\frac{1}{P_{m+1}^2}+\frac{2 S_{m+1}}{P_{m+2}}-\frac{y_0}{P_{m+2}} &= 
\frac{1}{y_{m+2}}\left(\frac{y_{m+2}}{P_{m+1}^2}+\frac{2}{P_{m+1}}\left(S_m+\frac{1}{P_{m+1}}\right)-\frac{y_0}{P_{m+1}}\right) \\&=
\frac{1}{y_{m+2}}\left(\frac{y_{m+2}+2}{P_{m+1}^2}-\frac{1}{P_m^2}\right) \\ &=
\frac{1}{y_{m+2}}\left(\frac{y_{m+1}^2}{P_{m+1}^2}-\frac{1}{P_m^2}\right) \\ &=
0
\end{align}
Here I used assumption in the second line.
If $y_0=2$, $\lim S_n =1$ by geometric sum. Obviously $y_0$ larger give smaller $\lim S_n $, so $S_n$ is bounded by $1$.Also, obviously $S_n$ is increasing. So $\lim S_n$ exist for all $y_0\geq2$. Let $L=\lim S_n$.
Take limit on $S_n^2-y_0 S_n+1=\frac{1}{P_n^2}$, we get $L^2-y_0L+1=0$. 
Solving the equation, we have $L=\frac{y_0\pm\sqrt{y_0^2-4}}{2}$
$L=\frac{y_0+\sqrt{y_0^2-4}}{2}$ is impossible for $y_0>2$ as it will larger than 1.
