If $$ \prod_{k=1}^n (x+\frac{1}{2^k}) = \sum_{k=0} ^n a_kx^k $$then find $$\lim_{n\to \infty}\frac{a_{n-2}}{a_{n-4}} = ?$$
I’m able to find $a_{n-2}$ easily but finding $a_{n-4}$ is quite difficult for me. I think involving binomial theorem or principle of inclusion and exclusion may work but still I’m not able to proceed with that.
It is simpler to recast this problem in terms of the function$$ f(x) = \frac{g(x)}{1+x} = g(x/2),\quad\text{where}\quad g(x)\doteq\prod_{k=0}^\infty \left(1+\frac{x}{2^k}\right). $$ The quantity you seek is the ratio of the coefficients of $x^2$ and $x^4$ in the power-series expansion of $f(x)$. But$$ f(x) = (-x/2;1/2)_\infty, $$ where $$ (a;q)_n \doteq \prod_{k=0}^{n-1} (1-a q^k) $$ denotes the $q$-Pochhammer symbol. Using the formula $$ (x;q)_\infty = \sum_{n=0}^\infty \frac{(-1)^n q^{n(n-1)/2}}{(q;q)_n} x^n, $$ we find that the power-series expansion of $f(x)$ is $$ f(x) = \sum_{n=0}^\infty \frac{2^{-n(n-1)/2} }{(1/2;1/2)_n} 2^{-n} x^n. $$ From the definition of $(a;q)_n$, we observe that $$ (1/2;1/2)_n = \frac{\alpha_n}{2^{n(n+1)/2}} $$ where $$ \alpha_n \doteq \prod_{i=1}^n (2^i-1) $$ is sequence A005329 in OEIS, with values of $1$, $1$, $3$, $21$, and $315$ for $n = 0$ to $4$. Substituting this into the above expansion for $f(x)$, we see that $$ f(x) = \sum_{n=0}^\infty \frac{x^n}{\alpha_n} = 1 + x + \frac{x^2}{3} + \frac{x^3}{3 \cdot 7} + \frac{x^4}{3 \cdot 7 \cdot 15} + \cdots. $$ Therefore the desired ratio is $(1/3) / (1/(3\cdot 7 \cdot 15)) = 7\cdot 15 = 105$.
Edit: The OP asked for a more elementary demonstration.
The first step is to realize that the way the problem is stated makes things artificially difficult by referencing coefficients whose indices are going to $\infty$. Therefore we divide each side of the original equation by $x^n$, substitute $1/x$ for $x$, and let $n \rightarrow \infty$ to arrive at the expression $$ f(x) \doteq \prod_{k=1}^\infty \left(1+\frac{x}{2^k}\right) =\sum_{n=0}^\infty c_n x^n. $$ Then the desired limit is $c_2/c_4$. Make sure you understand this transformation into a question about $f(x)$ before proceeding.
Now we need a formula for $c_n$. Trivially $c_0 = 1$. To find the other $c_n$ we use the identity $$ (1+x) f(x) = f(2x). $$ That is, multiplying $f(x)$ by $(1+x)$ is the same as changing the definition of $f(x)$ to begin with $k=0$. However, computing $f(2x)$ results in the same expression as $(1+x)f(x)$: we just have to relabel the indices by shifting $k$ by $1$. Make sure you understand how to prove the above identity before proceeding.
To find the coefficients $c_n$, we expand the identity above as $$ (1 + x) (c_0 + c_1 x + c_2 x^2 + \cdots) = c_0 + 2 c_1 x + 4 c_2 x^2 + \cdots. $$ Equating coefficients of $x$ in this expression yields the equalities $$ \begin{align} c_0 &= c_0,\\ c_0 + c_1 &= 2 c_1, \\ c_1 + c_2 &= 4 c_2, \\ c_2 + c_3 &= 8 c_3, \\ &\text{etc.} \end{align} $$ Simplifying these equations, we find that $$ c_n = \frac{c_{n-1}}{2^n-1}, $$ which, in conjunction with the initial value $c_0 = 1$, yields the required formula for the coefficients $c_n$.