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 $A\cong A^{op}$?



Serre functor on the category $Perf(A)$, $A$ - k-algebra
The assumption on the finite global dimension is just needed to have $Perf(A)=D^b(A-mod),$ which does not hold more generality.


It is indeed true that $Perf(A)$ has a Serre functor (given in the same way) when $A$ just is a Gorenstein algebra, that is the regular module $A$ has finite injective dimension as a left and right $A$-modules.


For example any selfinjective non-semisimple algebra (such as $K[x]/(x^n)$ for $n \geq 2$) is a Gorenstein algebra of infinite global dimension.


Good references are:
"Representation Theory: A Homological Algebra Point of View" by Zimmermann and the book by Happel on Triangulated Categories.


