= Solution
A precise version of the <Hilbert-Serre theorem> is as follows. Let $R=\bigoplus_{j\ge0}R_j$ be a commutative <graded ring>, generated over an <Artinian> degree-zero ring $R_0$ by homogeneous elements $y_1,\ldots,y_s$ of positive degrees $d_1,\ldots,d_s$. Let $M$ be a finitely generated <graded module>, bounded below. Each component has finite <module length> over $R_0$. Its <Poincare series of a graded module>, or <Hilbert series>, is
$$
P_M(t)=\sum_j\ell_{R_0}(M_j)t^j.
$$
When $R_0$ is a field, these lengths are dimensions. The theorem asserts
$$
\boxed{P_M(t)=\frac{q(t)}{\prod_{i=1}^s(1-t^{d_i})},\qquad q(t)\in\mathbb Z[t,t^{-1}].}
$$
For a nonnegatively graded module the numerator can be taken in $\mathbb Z[t]$. The finite-generation and finite-length hypotheses specify the meaning of the series; they are part of the theorem's statement.
We prove it by induction on $s$. Regard $M$ as a finite graded module over $S=R_0[Y_1,\ldots,Y_s]$, with $\deg Y_i=d_i$, acting through the specified generators of $R$. This polynomial ring is <Noetherian> by the <Hilbert basis theorem> and Question 1. When $s=0$, finitely many homogeneous generators over $R_0$ mean that only finitely many components are nonzero, so the series is a Laurent polynomial. For the induction step, put $y=Y_s$, $d=d_s$, $K=(0:_M y)$ and $C=M/yM$. Both $K$ and $C$ are finite graded modules; $y$ annihilates them, so they are modules over $R_0[Y_1,\ldots,Y_{s-1}]$. The finite generation of $K$ uses Noetherianity of $S$.
The <exact sequence> of graded modules
$$
0\longrightarrow K(-d)\longrightarrow M(-d)
\xrightarrow{\ y\ }M\longrightarrow C\longrightarrow0
$$
gives, by additivity of component lengths,
$$
(1-t^d)P_M(t)=P_C(t)-t^dP_K(t).
$$
Here the shift is defined so that $P_{M(-d)}=t^dP_M$. By induction both series on the right have denominator $\prod_{i<s}(1-t^{d_i})$. Dividing by $1-t^d$ proves the result. The annihilator term cannot be omitted unless multiplication by $y$ is injective.
In the standard grading $d_i=1$, the coefficients of $(1-t)^{-s}$ are $\binom{j+s-1}{s-1}$. Multiplying by the finite numerator shows that the component dimensions agree with a polynomial for sufficiently large $j$. Cumulative dimensions have generating function $P_M(t)/(1-t)$ and are also eventually polynomial. This consequence supplies the dimension used for <Weyl algebra> modules.
Use the <Bernstein filtration> $B_mA_n$, in which both coordinate generators and differentiation generators have degree one. The <ordered monomial basis of a Weyl algebra> gives
$$
\operatorname{gr}_B A_n\cong\mathbb C[x_1,\ldots,x_n,\xi_1,\ldots,\xi_n],\qquad
\dim_\mathbb C B_mA_n=\binom{m+2n}{2n}.
$$
Choose a finite-dimensional generating subspace $M_0$ of a nonzero finitely generated left module $M$ and set $M_m=B_mA_nM_0$. This is a <good filtration of a module>: the symbols of the generators generate its <associated graded module> over the displayed polynomial ring. The <Hilbert-Serre theorem> makes $\dim_\mathbb C M_m$ an eventual polynomial. Define the <Bernstein growth dimension of a Weyl algebra module> by
$$
\boxed{d(M)=\deg\bigl(\dim_\mathbb C M_m\text{ for }m\gg0\bigr).}
$$
The dimension is independent of the chosen finite generating subspace. Each of two finite generating sets is contained in a fixed filtration step for the other, giving $M_m\subseteq M'_{m+c}$ and $M'_m\subseteq M_{m+c'}$. Such fixed shifts preserve the degree of growth. The same argument applies to any good Bernstein filtration. This dimension agrees with the <Gelfand–Kirillov dimension of a module>; one may set $d(0)=-\infty$.
Let the nonconstant element $D$ have Bernstein degree $d\ge1$. Since $\operatorname{gr}_B A_n$ is a domain, the product of two nonzero symbols cannot vanish. Hence
$$
\deg_B(aD)=\deg_B a+d\quad(a\ne0),\qquad
(A_nD)\cap B_mA_n=(B_{m-d}A_n)D.
$$
Right multiplication by $D$ is injective, using the domain property of $A_n$. Thus the quotient filtration on $M=A_n/A_nD$ has cumulative dimensions
$$
\dim_\mathbb C M_m=
\binom{m+2n}{2n}-\binom{m-d+2n}{2n}\quad(m\ge d).
$$
The leading terms of degree $2n$ cancel, and the leading surviving term is
$$
\frac{d}{(2n-1)!}\,m^{2n-1}.
$$
It is nonzero, because $d\ge1$. Therefore
$$
\boxed{d(A_n/A_nD)=2n-1.}
$$
Equivalently the <Hilbert series> of the associated graded quotient is $(1-t^d)/(1-t)^{2n}$. Here $d$ is Bernstein degree, not differential order: multiplication by $x_1$ has differential order zero but Bernstein degree one. Using the order filtration in this dimension argument would lose the finite-dimensional filtration pieces.
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