Product of sums

Interpret $f(x)$ as a generating function. The coefficient of $x^p$ counts the number of ways to partition $p$ such that each part is in the form $A^k$ for some non-negative integer $k$ , and no part appears more than $A-1$ times. (To see this, observe that $\sum x^{nA^k}$ gives $n$ parts, all equal to $A^k$ .)

For any integer $p$ , there is exactly one such partition: interpret $p$ in base $A$ . For example, for $p = 7, A = 3$ , we have $p = 21_3$ , or $p = 2 \cdot 3^1 + 1 \cdot 3^0$ . Thus the required partition is $3^1 + 3^1 + 3^0$ . It's easy to see that there is no other such partition. Suppose there is one such partition besides the one obtained through base $A$ . Consider the largest difference; that is, the largest $A^k$ that appears in one partition and not in the other. The sum of all smaller parts cannot be equal to $A^k$ ; it's at most $(A-1)(A^{k-1} + A^{k-2} + \ldots + A^0) = A^k - 1$ . Thus the partition with $A^k$ must have larger sum; in particular, they don't sum to the same number, thus one of them is invalid. Since the base $A$ one is a valid partition, our assumed partition must be the invalid one.

Now that we know there is one such partition for each $p$ , we can rewrite $f(x)$ in a more standard generating function form: $\sum_{p=0}^\infty x^p$ , because the coefficient of $x^p$ counts the number of such partitions of $p$ , which is 1 for all $p$ . Thus $f(x) = \sum_{p=0}^\infty x^p = \frac{1}{1-x}$ with radius of convergence 1. Since $\left| \frac{1}{2} \right| < 1$ , $f \left( \frac{1}{2} \right)$ converges, and we can compute its value with that formula: $\frac{1}{1 - \frac{1}{2}} = \boxed{2}$ .

f\left( x \right) =\lim _{ n\longrightarrow \infty }{ \prod _{ k=0 }^{ n }{ \left( \sum _{ n=0 }^{ A-1 }{ { x }^{ n{ A }^{ k } } } \right) } =\quad } \\ \lim _{ n\longrightarrow \infty }{ \prod _{ k=0 }^{ n }{ \frac { { x }^{ { A }^{ k+1 } }-1 }{ { x }^{ { A }^{ k } }-1 } } } =\quad \lim _{ n\longrightarrow \infty }{ \frac { { x }^{ { A }^{ n+1 } }-1 }{ { x }-1 } } =\boxed { \frac { 1 }{ 1-x } } \quad \quad \quad \left\{ \left| x \right| <1 \right \\ \\ \therefore \quad f\left( \frac { 1 }{ 2 } \right) =\quad \boxed { 2 }

Similar solution with @Aditya Dhawan 's presented as follows.

$\begin{aligned} f(x) & = \prod_{k=0}^\infty \color{#3D99F6} \sum_{n=0}^{A-1} x^{nA^k} & \small \color{#3D99F6} \text{Sum of a GP} \\ & = \prod_{k=0}^\infty \color{#3D99F6} \frac {1-x^{A^{k+1}}}{1-x^{A^k}} & \small \color{#3D99F6} \text{for } |x| < 1 \\ & = \frac {\prod_{k={\color{#D61F06}1}} ^\infty (1-x^{A^k})}{\prod_{k={\color{#D61F06}0}}^\infty (1-x^{A^k})} \\ & = \frac 1{1-x^{A^{\color{#D61F06}0}}} \\ & = \frac 1{1-x} \end{aligned}$

$\implies f \left(\frac 12\right) = \dfrac 1{1-\frac 12} = \boxed{2}$ .