I have the next definition: $Tr(A)=\sum_n<u_n,Au_n>$, where $A$ is a positive linear operator on $H$ (Hilbert), and $\{u_n\}$ is an orthonormal base of $H$. And an operator is trace class if $Tr(A)<\infty$.
Let $A:D(A)\subset H\to H$ a positive, self-adjoint, densely defined operator in $H$, Hilbert. $A$ only has point spectrum $\sigma (A)=\{ \lambda_k\}_{k\in \mathbb{N}}$ and $\lambda_k < \lambda_{k+1}$ $\forall k$. $\{u_k\}_{k\in \mathbb{N}}$ is the respective orthonormal base of eigenvectors of $A$.
I know that the trace is independent of the base, its ciclicity, and that $<v,Av>\geq\lambda_1\|v\|$.
I got that $Tr(e^{-tA})=\sum_k e^{-t\lambda_k}$ but I don't know how to show it's finite if is possible.
Any help is regarded.
Consider the operator \begin{align} A = \sum^\infty_{n=1} \log(1+n) |u_n\rangle \langle u_n| \end{align} then we see that \begin{align} e^{-tA} = \sum^\infty_{n=1}e^{-t \log(1+n)}|u_n\rangle \langle u_n|. \end{align}
Note that \begin{align} \operatorname{Tr}(e^{-tA}) = \sum^\infty_{n=1} \frac{1}{(1+n)^t}. \end{align} which does not converge for $t\le 1$.