Evaluate $\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac{2}{45^{2^k}+45^{-2^k}}\right)$ Evaluate $$P=\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac{2}{45^{2^k}+45^{-2^k}}\right)$$
My try:
Let $a_k=45^{2^k}$ Then we have $45^{-2^k}=\frac{1}{a_k}$ 
So We get:
$$1+\frac{2}{a_k+\frac{1}{a_k}}=\frac{(a_k+1)^2}{a_k^2+1}$$
So
$$P=\lim_{n \to \infty}\prod_{k=0}^n\frac{(a_k+1)^2}{a_k^2+1}$$
Any way from here

Note that $a_k^2=a_{k+1}$, so 
$$
\prod_{k=0}^{n}\frac{(a_k+1)^2}{a_k^2+1}=\prod_{k=0}^{n}\frac{(a_k+1)^2}{a_{k+1}+1}
=\frac{a_0+1}{a_{n+1}+1}\cdot(a_0+1)(a_1+1)\ldots(a_n+1).
$$
Then, 
$$
a_k+1=\frac{a_k^2-1}{a_k-1}=\frac{a_{k+1}-1}{a_k-1},~\text{so}~
$$
$$
\prod_{k=0}^{n}\frac{(a_k+1)^2}{a_k^2+1}=\frac{a_0+1}{a_{n+1}+1}\cdot(a_0+1)(a_1+1)\ldots(a_n+1)=\frac{a_0+1}{a_{n+1}+1}\cdot\frac{a_{n+1}-1}{a_0-1}.
$$
Since $a_{n+1}\rightarrow +\infty$ when $n\rightarrow\infty$ we have 
$$
\lim_{n\rightarrow\infty}\prod_{k=0}^{n}\frac{(a_k+1)^2}{a_k^2+1}=\frac{a_0+1}{a_0-1}=\frac{45+1}{45-1}=\frac{23}{22}.
$$
