Consider a finite-dimensional $k$-algebra $A$ of finite global dimension. Then it is known that the Serre functor on $D^b(mod-A)$ exists and is given by the Nakayama functor. The proof goes something like this:

The $k$-duality $(-)^*=R\underline{\text{Hom}}_k(-,k)$ gives an equivalence $D^b(mod-A)\to D^b(A-mod).$ The $A$-duality $(-)^\vee=R\underline{\text{Hom}}_A(-,A)$ maps the category of perfect complexes $Perf(A)$ to $Perf(A^{op})$, because $A^\vee$ is isomorphic to $A$ in $A-mod$. Then one proves that for $M\in Perf(A)$, $N\in D^b(mod-A)$ we have

$$Hom_{D(Mod-A)} (M,N)^*\cong Hom_{D(Mod-A)} (N,M^{\vee*}).$$

Finally, we notice that for finite-dimensional algebras of finite global dimension $Perf(A)=D^b(A-mod).$

But what if $\text{r.gl.dim}(A)=\infty$? The only reason I see why the above proof might not give us a Serre functor on $Perf(A)$ is that $(-)^*$ may not map $Perf(A)$ to $Perf(A^{op})$, since $A$ is not necessarily isomorphic to $A^*$ as a left $A$-module. In this case, is it enough to have $A\cong A^{op}$?