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?" , but I don't know much about this subject. Thanks!)