Studying about the way Euler answered the Basel Problem, I was interested about functions defined by infinite products. So I developed a way to expand a product by the "general" form of the factors. For example, pick $f(x)=\prod_{n=1}^\infty\left(1+\frac{x}{2^n}\right)=(1+x/2)(1+x/4)(1+x/8)(1+x/16)\dots$
First term will be 1, since we would multiply all 1's of the factors of $f(x)$.
Second term we will make $x$, so we will have $\frac{x}{2}*\overbrace{1*1*1*...}^{\text{all the 1's from the 1nd, 3rd,... factors}}+\frac{x}{4}*\overbrace{1*1*1*...}^{\text{all the 1's from the 1nd, 2rd,4th, ... factors}}+...=x(\frac{1}{2}+\frac{1}{4}+...)=x\sum_{n=1}^\infty \frac{1}{2^{n}}$
Third term we will make $x^2$, so we need to pick 2 $x$'s since $(x/2^n) * (x/2^k) = x^2/2^{ n+k}$. So we need to pick the x from the first factor and multiply it to all others x from the other factors, like $x/2(x/4+x/8+...)=\frac{x^2}{2}\sum_{n=2}^\infty \frac{1}{2^n}$. Then you do the same to the 2nd x ($x/2$), but if you multiply it to the first x, you will repeat a combination, so the final result of that part is $x/4(x/8+...)$. The upcoming combinations will be the same, and in the final we will reach this: $$x^2\sum_{n=1}^\infty\sum_{k=n+1}^\infty \frac{1}{2^n2^k}$$
In fact, we can conclude that the (n+1)'th term is $$x^n\sum_{k_1\geq 1 \\ k_n>k_{n-1}>k_{n-2}>\dots>k_1}\prod_{j=1}^n \frac{1}{2^{k_j}}$$
If we want to change the coefficients of x of the factors of $f$, just substitute $\frac{1}{2^n} \text{to} \space a_n$, then you will have $f(x)=\prod_{n=1}^\infty (1+a_nx)$
Since the general form of the coefficients of this expansion is too complex for my small brain, we can think of making some functions of this type:
Let $B_0(m)=1$, and $B_{n+1}(m)=\sum_{k=m+1}^\infty a_kB_n(k)$. Thus, the (n+1)'th term is $x^nB_n(0)$. So $f(x)=\sum_{n=0}^\infty x^nB_n(0)$, which is the Maclaurin series of $f$. So $B_n(0)=\frac{1}{n!}\frac{d^nf}{dx^n}|_{x=0}$.
In the example I picked ($a_n=A^{-n}, \text{generalizing for all numbers}$), we have that $$f(x)=\sum_{n=0}^\infty x^n \prod_{k=1}^n \frac{1}{A^k-1}=\sum_{n=0}^\infty \frac{x^n}{2^{n(n+1)/2}}\prod_{k=1}^n \frac{1}{1-\frac{1}{A^k}}$$.
Can we make a general form of $B_n(0)$ for all $a_n$?