A few conjectured limits of products involving the Thue–Morse sequence (related to my previous questions$^{[1]}$$\!^{[2]}$)
Let's define the signed Thue–Morse sequence $t_n$ by the recurrence
$$t_0 = 1, \quad t_n = (-1)^n \, t_{\lfloor n/2\rfloor},\tag1$$
or by the generating function
$$\sum_{n=0}^\infty t_n \, x^n=\prod_{n=0}^\infty\left(1-x^{2^n}\right).\tag{$1^\prime$}$$
It seems that the following conjectures hold:
$$\lim_{n\to\infty}\prod_{k=0}^{2^n-1}\left(k+\tfrac12\right)^{t_k}\stackrel{\color{gray}?}=\frac12\tag2$$
$$\lim_{n\to\infty}\prod_{k=0}^{2^n-1}\left(k+1\right)^{t_k}\stackrel{\color{gray}?}=\frac1{\sqrt2}\tag3$$
$$\lim_{n\to\infty}\prod_{k=0}^{2^n-1}\left(k+1\right)^{(-1)^k\,t_k}\stackrel{\color{gray}?}=\frac1{2\sqrt2}\tag4$$
How can we prove these? Are there any other limits of products similar to these?
 A: 2) For $n\ge 1$ we have 
$$\prod_{k=0}^{2^n-1}\left(k+\tfrac12\right)^{t_k}=$$
$$\prod_{k=0}^{2^{n-1}-1}\left(2k+\tfrac12\right)^{t_{2k}}\left(2k+1+\tfrac12\right)^{t_{2k+1}}=$$
$$\prod_{k=0}^{2^{n-1}-1}\left(2k+\tfrac12\right)^{t_k}\left(2k+1+\tfrac12\right)^{-t_k}=$$
$$\prod_{k=0}^{2^{n-1}-1}\left(\frac{2k+\tfrac12}{2k+1+\tfrac12}\right)^{t_k}=$$
$$\prod_{k=0}^{2^{n-1}-1}\left(\frac{k+\tfrac14}{k+\tfrac34}\right)^{t_k}.$$
So, according to the introduction and Lemma 1 from the last paper you referenced (“Infinite products involving binary digit sums” by Samin Riasat), the left hand side of (2) equals 
$$\prod_{k=0}^{\infty}\left(\frac{k+\tfrac14}{k+\tfrac34}\right)^{t_k}=$$
$$\frac13\prod_{k=1}^{\infty}\left(\frac{k+\tfrac14}{k+\tfrac34}\right)^{t_k}=$$
$$\frac13f\left(\frac14,\frac34\right)=\frac13\cdot\frac32=\frac12.$$
3) Similarly to the previous case we can show that the left hand side of (3) equals $\tfrac12 f\left(\tfrac 12,1\right)=\tfrac 1{\sqrt2}$.
1) Here preliminary calculations are a bit longer. For $n\ge 2$ we have 
$$\prod_{k=0}^{2^n-1}(k+1)^{(-1)^kt_k}=$$
$$\prod_{k=0}^{2^{n-1}-1}(2k+1)^{t_{2k}}(2k+1+1)^{-t_{2k+1}}=$$
$$\prod_{k=0}^{2^{n-1}-1}(2k+1)^{t_k}(2k+2)^{t_k}=$$
$$\prod_{k=0}^{2^{n-2}-1}((4k+1)(4k+2))^{t_{2k}}((4k+3)(4k+4))^{t_{2k+1}}=$$
$$\prod_{k=0}^{2^{n-2}-1}((4k+1)(4k+2))^{t_k}((4k+3)(4k+4))^{-t_{k}}=$$
$$\prod_{k=0}^{2^{n-2}-1}\left(\frac{k+\tfrac14}{k+\tfrac34}\right)^{t_k}\left(\frac{k+\tfrac12}{k+1 }\right)^{t_k}.$$
Thus the left hand side of (1) equals the product of the left hand sides of (2) and (3), which is $\tfrac 1{2\sqrt2}.$
