Limit of a countable separable is countable separable?

Question

Let $\rho$ be a positive trace class operator on $H\otimes H$, where $H$ is a separable Hilbert space (not necessarily finite dimensional). We say that $\rho$ is countable separable if $\rho=\sum_{i=1}^{\infty}\lambda_i\alpha_i\otimes\beta_i $, where $\alpha_i,\beta_i$ are states ( positive trace class operators with trace equal to $1$ ) on $H$ and $\lambda_i\geq 0$ such that $\sum_{i=1}^{\infty}\lambda_i< \infty$.
Let $\{e_n\}$ be an orthonormal basis for $H$ and let $P_n$ be the orthogonal projection on to the space of span $\{e_i:1\leq i\leq n\}$. Are the following two statements equivalent?

1. $(I_H\otimes P_n)\rho (I_H\otimes P_n^*)$ is countable separable, for all $n$.
2. $\rho$ is countable separable.

Clearly, $2 \implies 1$. What about the converse?

Answer 1

Your "countable separability" is called "countable decomposability" in [1]. As shown there, there exist separable but not countably decomposable states (in fact such states are dense as it is conjectured in [1]). Since every separable state on the tensor product of finite dimensional Hilbert spaces is finitely decomposable, your condition 1 does not imply 2.

See the MO post Is the set of separable quantum states closed? for some relevant discussion (and my goof).

[1] A. S. Holevo, M. E. Shirokov, R. F. Werner; Separability and Entanglement-Breaking in Infinite Dimensions. (https://arxiv.org/abs/quant-ph/0504204). See also (https://iopscience.iop.org/article/10.1070/RM2005v060n02ABEH000830).

Answer 2

This need not be true in general as Echo pointed out, however if the coefficients $\lambda_{i,j}$ where i is for the operator and j is for the summation satisfies the requirements for monotone convergence, then the limit exists, specifically if you think of them as entries of a Matrix, the columns must be monotone and bounded and for each row the summation must converge. So all we have to guarantee is that the coefficients for the operators are bounded this can be done by for example requireing $tr(\rho_n) ≤ M\forall n$.