Let $I\subset \prod_n B(H_{m_n})$ be a separable $C^*$ algebra,where $\prod B(H_{m_n})$ denotes the $\ell ^{\infty}$direct sum of $B(H_n)$ and $dim(H_{m_n})<\infty$. We suppose that there exits a strictly positive element $A=(A_1,\cdots,A_n,\cdots)$ in $I$ (each $A_i \in B(H_{m_i})$) such that $\lim tr_{m_n}(A_n)=0$ and $A^\frac{1}{k},k=1,2,\cdots$ is an approximate unit for $I$.
Let $\{a_{1,n},\cdots,a_{m_n,n}\}$ denote the eigenvaluses of $A_n$,then $b_n=\{j:a_{j,n}\leq2tr_n(A_n)\}$ has cardiniality at least $\frac{m_n}{2}$, If $P=(P_1,\cdots,P_n,\cdots)$ is the orthogonal projection onto the span of the eigenvectors associated with $\{a_{jn}:j\in b_n\}$then we can conculde that $0\leq P_nA_n\leq 2 tr_n(A_n)P_n$and hence$\|P_nA_n\|\to 0$.
Can we use the fact that $A^\frac{1}{k},k=1,2,\cdots$ is an approximate identity element for $I$ to have the conclusion: For any $B=(B_1,\cdots,B_n,\cdots)\in I$,$\|P_nB_n\|\to 0$
I think nothing prevents you from taking $B_n=P_n$, so $\|B_nP_n\|=1$.