I'm currently working through Eisenbud's Geometry of Syzygies, specifically the following (partial)
Exercise 4.9. Say $M$ is a f.g. graded Cohen-Macaulay module over the graded $S=\mathbb{K}[x_0,\ldots,x_r]$, $y_0,\ldots,y_s$ is a maximal $M$-regular sequence of linear forms, and $M'=M/(y_0,\ldots,y_s)$. Then we have the following formula of Hilbert functions: $$H_M(d)=\sum_{e\leq d}\binom{d+\dim M}{\dim M}H_{M'}(d-e).$$
What if $M=M'$, so $\dim M=0$? Then for $d\geq 0$ don't we get that $H_M(d)$ is the sum of $H_m(d')$ for all $d'\leq d$? Surely that's not right. Is this a typo? If so, what should the statement be? Certainly if $M$ gives us a set of points then $\dim M=0$ and it should eventually be a positive constant?
I think you are right. The first formula he gives in terms of generating functions seems correct: $$\Psi_M(t) = \frac{\Psi_{M'}(t)}{(1 - t)^{\dim M}},$$ where $\Psi_M(t) = \sum_{d = 0}^\infty H_d(M) t^d$. The power series expansion of $1/(1- t)^{\dim M}$ is $$(1 - t)^{-\dim M} = \sum_{d = 0}^\infty {\dim M + d - 1 \choose d} t^d = \sum_{d = 0}^\infty {\dim M + d - 1 \choose \dim M - 1} t^d,$$ where we take the conventions ${-1 \choose 0} = {-1 \choose -1} = 1$, ${-1 \choose d} = 0$ for any $d > 0$, and ${n \choose -1} = 0$ if $n \geq 0$. Therefore, convolving the power series $\Psi_{M'}(t)$ and $(1 - t)^{-\dim M}$, the correct formula is $$H_M(d) = \sum_{e \leq d} {\dim M + e - 1 \choose \dim M - 1} H_{M'}(d - e).$$
In the $\dim M = 0$ case, the only nonzero term on the RHS occurs when $e = 0$, so the formula degenerates to the equality $H_M(d) = H_{M'}(d)$.