Let $\Omega \subset \mathbb{R}^n$ be bounded and consider the Laplace operator on $L_2 \left( \Omega \right)$ with domain $\mathrm{dom} \left( \Delta \right) = \left \{ u \in H_0^1 \left( \Omega \right) : \Delta u \in L_2 \left( \Omega \right) \right \}$. Then, $\Delta$ is self-adjoint and has a orthonormal basis of eigenvectors $\phi_n$ with eigenvalues
\begin{equation} \Delta \phi_n = \lambda_n \phi_n \end{equation}
where the sequence $\left( \left \lvert \lambda_n \right \rvert \right)_{n \in \mathbb{N}}$ increases monotonically.
I can now try to define the operator
\begin{equation} A = \sum\limits_{n = 0}^\infty a_n \Delta^n \end{equation}
for suitable $a_n \in \mathbb{R}$ on the maximally invariant subset of $\Delta$. Now, I would like to calculate $\mathrm{Tr} \left[ A^{-1} \right]$. Is this a well-posed problem? If done naively, I obtain
\begin{equation} \mathrm{Tr} \left[ A^{-1} \right] = \sum\limits_{m=0}^\infty \left \langle \phi_m, \left( \sum\limits_{n = 0}^\infty a_n \Delta^n \right)^{-1} \phi_m \right \rangle = \sum\limits_{m=0}^\infty \left( \sum\limits_{n = 0}^\infty a_n \lambda_m^n \right)^{-1} \end{equation}
But then, there exists a smallest $M \in \mathbb{N}$ such that for all $m \in \mathbb{N}_{\ge M}$
\begin{equation} \left \lvert \sum\limits_{n = 0}^\infty a_n \lambda_m^n \right \rvert = \infty \end{equation}
suggesting that my result is
\begin{equation} \mathrm{Tr} \left[ A^{-1} \right] = \sum\limits_{m=0}^{M-1} \left( \sum\limits_{n = 0}^\infty a_n \lambda_m^n \right)^{-1} \end{equation}
which looks surprisingly simple. Is this procedure valid, or can it least be made valid for some "nice" $a_n$?