Given a sequence $(a_n)_{n \ge 1}$ such that :
$$a_1>1$$ $$a_{n+1}=2a_n^2-1$$
prove that the infinite product
$$P_n:=\displaystyle\prod_{k=1}^n \left(1+\dfrac{1}{a_k}\right)$$
converges to
$$\sqrt{\dfrac{a_1+1}{a_1-1}}$$
My try :
the recurrence relation along with the fact that $a_1>1$ (yielding $a_n>1$) suggests introducing
$$a_n = \cosh(b_n)$$
let's first re-write the product $P_n$ as
$$P_n=2^{n-1}(1+a_1)\times \dfrac{Q_n}{a_n}$$
where
$$Q_n:= \displaystyle\prod_{k=1}^{n-1} a_k$$ ($n\ge 2$)(with usual product convention $Q_1:=1$)
Now the recurrence relation gives for positive natural $k$
$$a_{k+1}=\cosh(2b_k)$$
then the injectivity of $\cosh$ over positives, along with the recurrence relation and finite descending induction gives for positive natural $n$
$$a_k=\cosh\left( \dfrac{b_n}{2^k} \right)$$ (for $k$ between $1$ and $n$)
$$Q_n:= \displaystyle\prod_{k=1}^{n-1} \cosh\left(b_n/2^k\right)$$
this product is known to be simplifiable when multiplying it by $\sinh\left(\dfrac{b_n}{2^{n-1}}\right)$ , turning into
$$Q_n=\dfrac{\sinh(b_n)}{2^{n-1}\sinh\left(\dfrac{b_n}{2^{n-1}}\right)}$$
Combining the previous facts, we get :
$$P_n=\dfrac{\sinh(b_n)}{\sinh\left(\dfrac{b_n}{2^{n-1}}\right)} \times \dfrac{1+a_1}{a_n}$$
Now it seems that we should find an equivalent of $a_n$, thing that I'm stuck with .
$(a_n)$ is strictly increasing thus bounded below by $a_1>1$ and candidates to be finite limits of it are $1$ and $-\dfrac 1 2$ thus it diverges to $+\infty$.
I can't continue from here.
Thanks for any advice.
You got pretty close to the answer. Note that $b_n = 2^{n-1} b_1$. Thus, \begin{align} P_n = \dfrac{\sinh(2^{n-1}b_1)}{\sinh\left(b_1\right)} \dfrac{a_1+1}{\cosh(2^{n-1}b_1)} \xrightarrow{n\to\infty} \frac{a_1+1}{\sinh(b_1)} = \frac{a_1+1}{\sqrt{a_1^2-1}} = \sqrt{\frac{a_1+1}{a_1-1}}. \end{align}