Let $a_k(x,\xi)$ be a family of symbols on $\mathbf{R}^d \times \mathbf{R}^d$, where $a_k$ has order $\alpha_k$, and $\lim_{k \to \infty} \alpha_k = -\infty$. Then for another symbol $a(x,\xi)$, we write
$$ a(x,\xi) \sim \sum_{k = 0}^\infty a_k(x,\xi) $$
if the order of the symbols $r_N(x,\xi) = a(x,\xi) - \sum_{k = 0}^N a_k(x,\xi)$ tends to $-\infty$ as $N \to \infty$.
Conversely, given a family of pseudodifferential operators $\{ T_k \}$, and another operator $T$, we write
$$ T \sim \sum_{k = 0}^\infty T_k $$
if for any $\alpha \in \mathbf{R}$, there exists $N_0$ such that for $N \geq N_0$, the operators $R_N = T - \sum_{k = 0}^N T_k$ can be expressed in terms of a symbol of order $\alpha$, i.e. there exists $a_N$ of order $\alpha$ such that for any smooth, compactly supported function $f$,
$$ R_N f(x) = \int a_N(x,\xi) e^{2 \pi i \xi \cdot (y-x)} f(y)\; dy. $$
Are these two notions `equivalent' in some sense, i.e. what can we say about a family of symbols $\{ a_k \}$ such that we have an asymptotic expansion
$$ T_a \sim \sum_{k = 0}^\infty T_{a_k}. $$
Is is then true that
$$ a \sim \sum_{k = 0}^\infty a_k\; \text{?} $$
If not, can we say something weaker?
