What is the infinite series corresponding to the infinite product below?
$$f(\alpha,x)=\prod_{n=1}^\infty \left(1+\frac{x}{\alpha^n}\right)$$
Edit:
Martin R told me to speak about what I have gotten up to now about how to expand this, so: up to now I only got really complicated expressions with double sums for example. I think this is related to partitions but I don't know exactly how. The reason for thinking this function is related to partitions is because at $x=1$ it is the generating function at $\frac{1}{\alpha}$ for the partition function, and at $\frac{1}{\alpha}$ it actually converges, as long as $\alpha>1$ (for $\alpha\in\mathbb{R}$
Edit 2:
I am looking for a clean expression like a power series with some coefficients I don't yet know. I wanted to know this because I was trying to evaluate something else that is related to partitions looking at it now.
Best Answer
The answer obviously is:
$$1+{x\over\alpha-1}+{x^2\over(\alpha-1)\cdot(\alpha^2-1)}+\dots+{x^n\over(\alpha-1)\cdot...\cdot(\alpha^n-1)}+\dots$$
How so?
Well, what is the coefficient at $x^n$? That's an infinite sum of various inverse powers of $\alpha$, some of them identical. How many terms $\frac1{\alpha^m}$ are there? As many as there are sets of $n$ different integers that sum to $m$. Is there any shorter way to express this? Why, sure; it is the same as in the problem "in how many ways can you give such-and-such sum in change, if you have such-and-such coins".
Taking the term at $x^2$ as an example:
$${1\over(\alpha-1)(\alpha^2-1)} = {1\over\alpha^3}\cdot\frac1{\left(1-{1\over\alpha}\right)\left(1-{1\over\alpha^2}\right)} = \\= \left({1\over\alpha}+{1\over\alpha^2}+{1\over\alpha^3}+\dots\right)\left({1\over\alpha^2}+{1\over\alpha^4}+{1\over\alpha^6}+\dots\right)= \\ = {1\over\alpha^3}+{1\over\alpha^4}+{2\over\alpha^5}+{2\over\alpha^6}+{3\over\alpha^7}+{3\over\alpha^8}+\dots$$ which is the same as I said earlier: the sum of inverse powers of $\alpha$ with certain combinatorial expression in the numerator.
So it goes.