Compute $ \lim_{n\to \infty}\prod_{i=1}^n B(p_i^{-2})$ 
Let $B(x) = \begin{pmatrix} 1 & x \\x & 1 \end{pmatrix}$, and $2=p_1<p_2<\cdots <p_n <\cdots$ primes number.
Compute $$\displaystyle \lim_{n\to \infty}\prod_{i=1}^n B(p_i^{-2})$$

I am sorry to post this but it's the first time I see an exercice about the convergence of infinite product of matrices. I do not know how can I deal.
Can someone enlighten me ?
 A: Result:
$$\boxed{\displaystyle\lim_{n\rightarrow\infty}\prod_{i=1}^n B\left(p_i^{-2}\right)=\frac{3}{2\pi^2}\left(\begin{array}{cc}
7 & 3 \\ 3 & 7\end{array}\right)}\tag{$\heartsuit$}$$

Derivation:
Let us write
$$B(x)=\sqrt{1-x^2}\,\exp\left\{ a(x)\left(\begin{array}{cc}
0 & 1 \\ 1 & 0\end{array}\right)\right\},$$
where $\displaystyle a(x)=\frac12\ln{\frac{1+x}{1-x}}$. Therefore the limit is equal to
$$\left[\prod_p\left(1-p^{-4}\right)\right]^{1/2}\exp\left\{ \frac12\left(\begin{array}{cc}
0 & 1 \\ 1 & 0\end{array}\right)
\ln\left(\prod_{p}\frac{1+p^{-2}}{1-p^{-2}}\right)\right\}.\tag{$\diamondsuit$}$$
Now using the products (see (10) and (11) here)
$$\prod_p\left(1-p^{-s}\right)=\frac{1}{\zeta(s)},\qquad 
\prod_p\left(1+p^{-s}\right)=\frac{\zeta(s)}{\zeta(2s)},$$
the formula ($\diamondsuit$) can be rewritten as
\begin{align}\frac{1}{\sqrt{\zeta(4)}}\exp\left\{ \left(\begin{array}{cc}
0 & 1 \\ 1 & 0\end{array}\right)
\ln\frac{\zeta(2)}{\sqrt{\zeta(4)}}\right\}=\frac{3\sqrt{10}}{\pi^2}\exp\left\{ \left(\begin{array}{cc}
0 & 1 \\ 1 & 0\end{array}\right)
\ln\sqrt{\frac{5}{2}}\right\}=\frac{3}{2\pi^2}\left(\begin{array}{cc}
7 & 3 \\ 3 & 7\end{array}\right).\end{align}
A: The other approach is to see, as you noted, that there is a single $P$ such that for all $x$:
$$P^{-1}B(x)P = \begin{pmatrix}1-x&0\\0&1+x\end{pmatrix}$$
Then $$\prod_{i=1}^n B(p_i^{-2}) = P\begin{pmatrix}\prod_1^n (1-p_i^{-2})&0\\0&\prod_1^n (1+p_i^{-2})\end{pmatrix}P^{-1}$$
So, since $\prod_1^\infty (1-p_i^{-2}) = 1/\zeta(2)=\frac{6}{\pi^2}$ and $\prod_1^\infty (1+p_i^{-2}) = \zeta(2)/\zeta(4)=\frac{15}{\pi^2}$, so this gives:
$$P\begin{pmatrix}\frac{6}{\pi^2}&0\\0&\frac{15}{\pi^2}\end{pmatrix}P^{-1}$$
Now, $P=\begin{pmatrix}1&1\\-1&1\end{pmatrix}$ and $P^{-1}=\frac{1}{2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}$. So we get:$$\frac{3}{2\pi^2}\begin{pmatrix}7&3\\3&7\end{pmatrix}$$
Which is the same result as above.
