Evaluate $\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac{2}{45^{2^k}+45^{-2^k}}\right)$

Question

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

Answer 1

Observe that

Hint:

$$(a_k+1)(a_k-1)=(45^{2^k})^2-1=a_{k+1}-1$$

$$(a_0+1)\prod_{k=1}^n(a_k+1)=a_{n+1}-1$$

Similarly, $$(a_0^2+1)\prod_{k=1}^n(a_k^2+1)=a_{n+1}^2-1$$

Answer 2

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 $$ \lim_{n\rightarrow\infty}\prod_{k=0}^{n}\frac{(a_k+1)^2}{a_k^2+1}=\frac{a_0+1}{a_0-1}=\frac{45+1}{45-1}=\frac{23}{22}. $$