What is... $$\lim_{\omega \to \infty} \left( {1 \over {a^{\omega}}} \cdot \prod_{N=1}^{\omega} (1+e^{b \cdot c^{-N}}) \right)$$
I'd like closed form solutions, and in this case that means any solution. So you don't need to concern yourself over "elementary" solutions. (although I'd prefer that). As I discuss below, the values that are infinite or zero are already known, I'd like expressions for finite values only...
What I know: I made a bounty on this question. I got a proof that showed that the limit of the above equation is infinite for $1 \lt a=c \lt 2$ and 0 for $a=c \gt 2$. So knowing the limit is volatile, I've made a generalization to find the analytical methods to get the finite values from this equation.
My attempt: I have absolutely no clue except for the case of $a=2 \ $, $b=1$ and $c=2$, which is just a horrible coincidence, considering that its the only finite non zero limit preserved in the proof given in the link, which does generalize to any b...
Create a line integral over the unit line evaluated with a uniform measure... $$\int_L e^x d \mu=\int_{L/2} e^x \ d\mu+\int_{L/2} e^{x+1/2} \ d\mu$$ This identity should be evident by self-similarity. Prepare for recursion... $$\int_L e^x d \mu=\int_{L/2} e^x+e^{x+1/2} \ d\mu=(1+e^{1/2}) \cdot \int_{L/2} e^x \ d\mu$$ $$\Rightarrow \int_L e^x d \mu=(1+e^{1/2}) \cdot \left( \int_{L/4} e^x \ d\mu+\int_{L/4} e^{x+1/4} \ d\mu \right)$$ $$\Rightarrow \int_L e^x d \mu=(1+e^{1/2}) \cdot (1+e^{1/4}) \cdot \left( \int_{L/4} e^x \ d\mu \right)$$ It wouldn't be hard to prove by induction then that... $$\Rightarrow \int_L e^x d \mu=\lim_{\omega \to \infty} \left(\prod_{N=1}^{\omega} (1+e^{2^{-N}}) \cdot \int_{L/{2^{\omega}}} e^x \ d\mu \right)$$ Yet we know what the left hand side equals, since it can be evaluated as a definite integral, also we know what the integral on the right equals. Since the measure is uniform and the number of values x will be allowed to take on the interval decreases to just the value, namely $0$... $$e-1= \lim_{\omega \to \infty} \left( {1 \over {2^{\omega}}} \cdot \prod_{N=1}^{\omega} (1+e^{2^{-N}}) \right)$$
Motivation: Getting an answer will allow me to derive methods to integrate a function like $e^x$ over fractals. As I hinted at above, its easy to find the numerical solution to the above limit. For instance for $a=2 \ $ , $b=2$ , $c=3$, the limit is $1.753$...
