Let $A$ be a Gorenstein algebra(We call $A$ Gorenstein provided that the regular module $A$ has finite injective dimension on both sides. And if $A$ is Gorenstein, then $inj.dim _AA= inj.dim A_A$ ), $A$-Gproj is the full subcategory of $A$-mod consisting of Gorenstein projective modules. We call $A$ $d$-Gorenstein if $inj.dim _AA= inj.dim A_A \leq d$. We call $A$ CM-finite provide that up to isomorphism there are only finitely many indecomposable modules in $A$-Gproj.
For each $d \geq 1$, denote by $\Omega^d(A-mod)$ the class of modules of the form $\Omega^d(M)$ for a module $M$. There is an equivalence: the algebra $A$ is $d$-Gorenstein iff $A$-Gproj =$\Omega^d(A-mod)$. Now how to use this equivalence get that Gorenstein algebra $A$ is CM-finite implies the algebra $A$ has finite global dimension?(By the equivalence, we can get $\Omega^d(A-mod)$ is representation-finite, then I don't know what to do next)
This is not true: For example, take $A=k[x]/(x^n)$. This is a self-injective algebra, so it is $0$-Gorenstein. Therefore, every $A$-module is Gorenstein-projective, and moreover there are only finitely many indecomposable $A$-modules (remember the classification of nilpotent endomorphisms of vector spaces). Nonetheless, $A$ is of infinite global dimension.