We say that $A\in B(\mathcal{H})$ is a trace class operator, if $\sum_{i\in I}\langle|A|e_i,e_i\rangle<\infty$,$\hspace{0.1cm}$ such that {$e_i; i\in I$} is a orthonormal bass for Hilbert space $\mathcal{H}$.
If compact operator on Hilbert space $\mathcal{H}$ is a trace class operator ?
No. For example, say $$Ax=\sum\lambda_n\langle x,e_n\rangle e_n,$$where $(e_n)$ are orthomormal and $\lambda_n$ are scalars. Then $A$ is compact if $\lambda_n\to0$, while $A$ is trace class requires $\sum|\lambda_n|<\infty$.
Unfortunately, the opposite direction is ture.$\renewcommand{\l}[1]{\lVert #1\rVert}\def \yb{\frac{1}{2}}$
Suppose $u$ is a positive trace class operator and $\{e_\alpha\}_{\alpha\in\Gamma}$ is a basis of $H$, then $\forall\varepsilon>0,$ there exists a finite subset $\Gamma_0$ of $\Gamma$ such that \begin{equation*} \lvert\sum_{\alpha\in \Gamma_0}\langle u e_\alpha,e_\alpha\rangle-\l{u}_1\rvert <\varepsilon, \end{equation*} i.e.$\newcommand{\Tr}{\operatorname{trace}}$ \begin{equation*} \lvert\Tr(p)-\Tr(u)\rvert<\varepsilon, \end{equation*} where \begin{equation*} p=\sum_{\alpha\in \Gamma_0}u^\yb e_\alpha\otimes e_\alpha u^\yb \end{equation*} is a finite-rank operator on $H$. Note that $\l{p-u}\leq\l{p-u}_1=\Tr(\lvert u-p\rvert)=\Tr(u-p), $ so $u\in K(H)$.
No. Compact, positive definite operators, for example, are trace class if and only if their eigenvalues are summable
