compute the Hilbert series of a free product of algebras We work over a field K. Assume there are two connected graded algebras $B, C$. And $A=B * C$.
The Hilbert series of A is
$A(x) = \sum_{n \geq 0} dim(A_n) x^n$
Why the Hilbert series satisfy the equation:
$\frac{1}{A(x)} = \frac{1}{B(x)} + \frac{1}{C(x)} -1$
This question comes from an article "Combinatorial Hopf algebras and generalized Dehn–Sommerville relations" (6.4). It says this is a well-known fact. But I don't know how to prove it.
 A: This is false without the hypothesis that $B$ and $C$ are connected, meaning that $B(0) = \dim B_0 = C(0) = \dim C_0 = 1$. With this hypothesis, the desired relation is equivalent to
$$\begin{align*} A(x) &= \frac{1}{\frac{1}{B(x)} + \frac{1}{C(x)} - 1} \\
 &= \frac{B(x) C(x)}{B(x) + C(x) - B(x) C(x)} \\ 
 &= \frac{B(x) C(x)}{1 - (B(x) - 1)(C(x) - 1)} \\
 &= B(x) C(x) \sum_{k=0}^{\infty} (B(x) - 1)^k (C(x) - 1)^k \end{align*}$$
which we can prove as follows: the free product consists of the direct sum of all tensor products of the form
$$B_{i_0} \otimes C_{j_0} \otimes B_{i_1} \otimes C_{j_1} \otimes \dots \otimes B_{i_{k+1}} \otimes C_{j_{k+1}}$$
where $i_0 \ge 0, j_{k+1} \ge 0$ and all the other indices are strictly positive (this is where we use the connectivity hypothesis; without it $\dim A_0$ would be infinite because $A_0$ itself would already be the free product of $B_0$ and $C_0$). This summand lives in degree $\sum_m (i_m + j_m)$ and for fixed $k$ such summands correspond exactly to terms in the product
$$B(x) \underbrace{ (C(x) - 1)(B(x) - 1)(C(x) - 1) \dots (B(x) - 1)}_{k \text{ pairs}} C(x)$$
where $B(x)$ corresponds to $i_0$, $C(x)$ corresponds to $j_k$, and the other factors correspond to the other indices in order.
As a sanity check we can take $B$ to be the free algebra on $m$ generators and $C$ to be the free algebra on $n$ generators (in degree $1$ in both cases), with Hilbert series $\frac{1}{1 - mx}, \frac{1}{1 - nx}$. Then $A$ is the free algebra on $m + n$ generators, with Hilbert series $\frac{1}{1 - (m+n)x}$, and we can easily verify that
$$1 - (m+n) x = (1 - mx) + (1 - nx) - 1.$$
