Let $0<t<1$ be a parameter. Let $n\in\mathbb{Z}$, $n>0$. For $i\in\{0,1,\dots,2^n-1\}$, we always consider $i$ as having $n$ binary digits (positions $1$ to $n$), putting $0$s if necessary. Let $f(i)$ be the number of times the $j+1$-th binary digit of $i$ is different from the $j$-th digit, $1\leq j\leq n-1$. Let $g(i)$ be the number of $1$s in the binary expansion of $i$. We consider
$$
p(t,n)=\frac{\sum\_{\substack{i=0\\i\equiv 1 \bmod 2}}^{2^n-1}(1/2)(1-t)^{n+f(i)-g(i)}(t)^{n-1-f(i)+g(i)}}{\sum\_{i=0}^{2^n-1}(1/2)(1-t)^{n+f(i)-g(i)}(t)^{n-1-f(i)+g(i)}}.
$$
Numerical evidence shows that $p(3/4,n) \rightarrow \sqrt{3}/2$ as $n\rightarrow \infty$. May it be possible to inquire on how to prove it? Thanks very much. 


(Perhaps this is similar to various limits involving binomial coefficients such as in the question "[A limit involving binomial coefficients?](https://mathoverflow.net/questions/87356/a-limit-involving-binomial-coefficients/)" , but I don't know much about this subject. Thanks!)



Limit of quotients of polynomials at fixed value
First, cancelling common factors we get
$$p(t,n) = \frac{ \sum\_{i=0\atop i\equiv 1\pmod{2}}^{2^n-1} \left(\frac{t}{1-t}\right)^{g(i)-f(i)} }{ \sum\_{i=0}^{2^n-1} \left(\frac{t}{1-t}\right)^{g(i)-f(i)} }$$


Clearly, $f(i)$ is the number of 2-mers "01" and "10", while $g(i)$ is the number of 2-mers $10$ and $11$ in the binary representation of $i$. Then $g(i)-f(i)$ is the number of $11$ minus the number of $01$.


Let $w = \frac{t}{1-t}$. Consider the [de Bruijn graph](https://en.wikipedia.org/wiki/De_Bruijn_graph) $G$ on the nodes $\{ 00, 01, 10, 11\}$, where all the arcs are weighted as follows:


* arcs incoming to $11$ have weight $w$;
* arcs incoming to $01$ have weight $w^{-1}$;
* the other arcs have weight 1.


That is, the adjacency matrix of $G$ is
$$A = \begin{bmatrix} 
1 & w^{-1} & 0 & 0 \\
0 & 0 & 1 & w \\
1 & w^{-1} & 0 & 0 \\
0 & 0 & 1 & w
\end{bmatrix}.
$$


We interpret each $n$-bit integer (prepended and appended by "0") as a walk of length $n$ in $G$ starting at nodes $00$ or $01$ and ending at nodes $00$ or $10$.


Then the numerator of $p(t,n)$ is the total weight of all walks of length $n$ from node $01$ to node $00$ or $10$, while the denominator of $p(t,n)$ is the total weight of all walks of length $n$ from node $00$ or $01$ to node $00$ or $10$. That is,
$$p(t,n) = \frac{[0,1,0,0]\cdot A^n\cdot [1,0,1,0]^T}{[1,1,0,0]\cdot A^n\cdot [1,0,1,0]^T}.$$


From here, one can get an explicit formula for $p(t,n)$, but for the purpose of the question it's enough to consider $t=\frac{3}{4}$ and $w=3$. It follows that
$$p(\frac{3}{4},n) = \frac{\sqrt{3}}{2}\cdot \left(1+O\left(\left(\frac{3-\sqrt{3}}{3+\sqrt{3}}\right)^n\right)\right).$$


