Let $(A,D(A))$ a closed, densely defined, unbounded linear operator on a Banach space $X$ to itself.
Consider the formal series $$ e^{tA}:= \sum_{n=0}^\infty t^n \frac{A^n}{n!} $$
and let $f \in D(A^\infty):=\bigcap_n D(A^n)$
By the Cauchy-Hadamard theorem if
$$ \rho:=\operatorname{limsup}_n \left(\frac{||A^n f||_X}{n!}\right)^{\frac{1}{n}}<\infty $$
$e^{tA}f$ converges absolutely in $X$ for $t \in \left[0,\frac{1}{\rho}\right)$. In particular, if $\rho=0$ It converges for any $t\ge0$.
I always see this theorem stated for bounded operators, but i can't see any reason for which the same proof shouldn't apply also for unbounded operators, assuming that $\rho$ is not infinity.
My question is: there is an example of an operator and a function on a Banach space for which the $\rho=\infty$. In particular, if $A=\Delta$ the laplacian on (the closure of) $C^\infty(\mathbb{R}^n)$ there is a function in $D(\Delta^\infty)$ for which such series doesn't converges for any $t$?