Self adjoint operators and trace class property

The series is required to converge for any basis, and so it has to survive reordering, and thus absolute convergence.

Since the real and imaginary parts of $T$ will satisfy the hypothesis and linear combinations of trace-class are trace-class, we may assume that $T$ is selfadjoint. We may also assume that $T$ is compact; because if $T$ is not compact, there exists $\lambda\in\sigma(T)\setminus\{0\}$ and $\delta>0$ such that $\lambda-\delta>0$ and the spectral projection $E_T(\lambda-\delta,\lambda+\delta)$ is infinite, and so an orthonormal basis of its range, extended to an orthonormal basis of $H$, provides $\{f_j\}$ such that $\sum_n|\langle Te_n,e_n\rangle|=\infty$.

Knowing that $T$ is compact, by the Spectral Theorem, we know that $$\tag1T=\sum_j\lambda_jP_j,$$ where $\lambda_j\in\mathbb R\setminus\{0\}$ for all $j$, and the projections $P_j$ are rank-one and pairwise orthogonal.

If $T$ is not trace-class, then $$ \operatorname{Tr}(T)=\sum_j|\lambda_j|=\infty. $$ If we write $\lambda_j^+$ for the positive eigenvalues and $\lambda_j^-$ for the negative ones, at least one of $\sum_j\lambda_j^+$ and $\sum_j\lambda_j^-$ diverges. Then, with $\{e_j\}$ the orthonormal basis given by $(1)$ (i.e., $P_j=\langle\cdot,e_j\rangle\,e_j$), we have that $$ \sum_j\langle Te_j,e_j\rangle $$ cannot converge absolutely.