Convergence in trace Let $A$ and $B$ be two self-adjoint, positive definite Compact operators on a Hilbert space $\mathcal{H}$. Further, let $A$ be trace class. Define $C_n \equiv AB(\frac{I}{n} + BAB)^{-1}$. Does $\frac{tr(C_nC_n^{T})}{n}$ converge to $0$ as $n \to \infty$. The answer is affirmative when $A$ and $B$ commute as can be found in section 7.1 of the following paper-
https://arxiv.org/pdf/1103.2692.pdf
 A: Yes, this is true. Denote $Q_n=\sqrt{n}\sqrt{A}B$. We have to prove that $c_n:=tr(\sqrt{A}Q_n(I+Q_n^*Q_n)^{-2}Q_n^*\sqrt{A})\to 0$. We have $c_n=tr(AQ_n(I+Q_n^*Q_n)^{-2}Q_n^*)$. We need a
Lemma. For any compact operator $Q$ we have $Q(I+Q^*Q)^{-2}Q^*\leqslant \frac14I$. Moreover, for any vector $x$ we have $\|Q_n(I+Q_n^*Q_n)^{-2}Q_n^* x\|\to 0$.
Proof. We may choose orthonormal bases $(v_i)$ and $(u_i)$ and positive numbers $\lambda_i$ so that $Qx=\sum \lambda_i (x,u_i)v_i$, $Q^*x=\sum \lambda_i (x,v_i)u_i$, then $(v_i)$ is eigenbasis for $Q(I+Q^*Q)^{-2}Q^*$ with eigenvalues $\lambda_i^2/(1+\lambda_i^2)^2\leqslant 1/4$. For $Q=Q_n$ bases $v_i,u_i$ are fixed, $\lambda_i=\sqrt{n}\mu_i$ for some fixed positive numbers $(\mu_i)$, thus each specific eigenvalue $n\mu_i^2/(1+n\mu_i^2)^2$ tends to 0. Representing $x$ as a sum of small vector and a finite linear combination of $v_i$ we get the second claim. 
Now let $e_i$ be an eigenbasis for $A$, $Ae_i=t_ie_i$, $\sum t_i<\infty$. We have $$c_n=\sum t_i (Q_n(I+Q_n^*Q_n)^{-2}Q_n^*e_i,e_i),$$
where $i$-th term always does not exceed $t_i$ and each specific term tends to 0. So we may simply for given $\varepsilon>0$ choose $N$ so that $\sum_{i>N} t_i<\varepsilon$, estimate the tail of (1) for $i>N$ as $\varepsilon/4$ and say that first $N$ summands are less than $\varepsilon/4$ provided that $n$ is large enough. Totally $|c_n|<\varepsilon/2$ for large $n$, as desired. 
