Evaluate the limit $\lim_{n \rightarrow \infty}\sum_{k=1}^n\frac{1}{x_{1}^{2}x_{2}^2...x_{k}^2}$ We have the sequences $(x_{n})_{n\geq1}$$(y_{n})_{n\geq1}$ with positive real numbers. $x_{1}=\sqrt{2}$, $y_{1}=1$ and $y_{n}=y_{n-1}\cdot x_{n}^{2}-3$ for every $n\geq2$. We know the sequence $(y_{n})_{n\geq1}$ is bounded. Find $$\lim_{n \rightarrow \infty}\sum_{k=1}^n\frac{1}{x_{1}^{2}x_{2}^2...x_{k}^2}$$
I tried rewriting $y_{n}$ only with terms from $(x_{n})_{n\geq1}$ but we get an ugly formula. I don't know exactly how we can we use the fact that $(y_{n})_{n\geq1}$ is bounded.

Introduce an auxilary sequence $z_n = \frac{y_n}{y_n+3} \iff y_n = 3\frac{z_n}{1-z_n}$. For $i \ge 2$, we have
$$y_i = y_{i-1}x_i^2 -3
\implies \frac{1}{x_i^2} = \frac{y_{i-1}}{y_i+3} = 
\frac{3\frac{z_{i-1}}{1-z_{i-1}}}{
3\frac{z_i}{1-z_i} + 3} = z_{i-1}\frac{1-z_i}{1-z_{i-1}}
$$
Multiply from $i = 2$ to any $k \ge 2$, we get
$$\prod_{i=2}^k \frac{1}{x_i^2} 
= \frac{1-z_k}{1-z_1}\prod_{i=1}^{k-1}z_i 
= \frac{1}{1-z_1}\left(\prod_{i=1}^{k-1}z_i - \prod_{i=1}^{k}z_i\right)
$$
Summing from $k = 2$ to $n$, we get
$$\sum_{k=2}^n \prod_{i=2}^k \frac{1}{x_k^2}
= \frac1{1-z_1}\left(z_1 - \prod_{i=1}^n z_i\right)
$$
This leads to
$$\sum_{k=1}^n \prod_{i=1}^k \frac{1}{x_k^2}
= \frac{1}{x_1^2}\left( 1 + \sum_{k=2}^n \prod_{i=2}^k \frac{1}{x_k^2}\right) = \frac1{x_1^2(1-z_1)}\left(1 - \prod_{i=1}^n z_i\right)$$
If $y_n$ is bounded from above by $M$, then $z_n$ is bounded from above by $\frac{M}{M+3} < 1$.
As a result, $\prod\limits_{i=1}^n z_i \le \left(\frac{M}{M+3}\right)^n$ and hence converges to $0$ as $n \to \infty$.
From this, we can deduce
$$\lim_{n\to\infty}
\sum_{k=1}^n \prod_{i=1}^k \frac{1}{x_k^2}
= \frac1{x_1^2(1-z_1)} = \frac{y_1+3}{3x_1^2} = \frac23$$
