[proofplan] We construct $a$ by a Borel-type cutoff summation: multiply the $j$th coefficient by a scalar cutoff $\chi(\varepsilon_j h)$ whose active range moves toward $h=0$ as $j$ grows. The parameters $\varepsilon_j \to \infty$ are chosen inductively so that every fixed symbol seminorm sees a summable tail after factoring out any prescribed power $h^N$. Local finiteness in $j$ makes the candidate a smooth symbol family, and the imposed seminorm estimates give the asymptotic expansion. Uniqueness follows immediately by subtracting two such sums and testing the defining estimate for every $N$. [/proofplan]

[step:Choose symbol seminorms and cutoff scales that make the tails summable]Let $\mathbb{N}_0 := \mathbb{N} \cup \{0\}$. For multi-indices $\alpha,\beta \in \mathbb{N}_0^n$, define the symbol seminorm \begin{align*} q_{\alpha,\beta}(b) := \sup_{(x,\xi,h) \in T^*\mathbb{R}^n \times (0,h_0]} \frac{|\partial_x^\alpha \partial_\xi^\beta b(x,\xi;h)|}{m(x,\xi)} \end{align*} for every smooth family \begin{align*} b: T^*\mathbb{R}^n \times (0,h_0] \to \mathbb{C}. \end{align*} Choose an enumeration $(q_k)_{k \ge 1}$ of the seminorms $q_{\alpha,\beta}$, with repetitions allowed if needed. Thus $b \in S(m)$ exactly when $q_k(b) < \infty$ for every $k \ge 1$. Choose a cutoff \begin{align*} \chi: \mathbb{R} \to \mathbb{R} \end{align*} with $\chi \in C_c^\infty(\mathbb{R})$, $\chi(t)=1$ for $|t| \le 1$, and $\chi(t)=0$ for $|t| \ge 2$. We choose a sequence $(\varepsilon_j)_{j \ge 0}$ in $(0,\infty)$ such that $\varepsilon_j \to \infty$ and, for every $j \ge 1$, every $0 \le N < j$, and every $1 \le k \le j$, \begin{align*} q_k\bigl(h^{-N} h^j \chi(\varepsilon_j h)a_j\bigr) \le 2^{-j}. \end{align*} This is possible because, on the support of $h \mapsto \chi(\varepsilon_j h)$, one has $0 < h \le 2/\varepsilon_j$. Since $a_j$ is $h$-independent and $\chi(\varepsilon_j h)$ has no $x$- or $\xi$-derivatives, for $0 \le N < j$, \begin{align*} q_k\bigl(h^{-N} h^j \chi(\varepsilon_j h)a_j\bigr) \le \|\chi\|_\infty q_k(a_j)\left(\frac{2}{\varepsilon_j}\right)^{j-N}. \end{align*} For fixed $j$, the indices $N<j$ and $k \le j$ form a finite set, so choosing $\varepsilon_j$ sufficiently large gives all the required inequalities. Increasing $\varepsilon_j$ further if necessary, we may also impose $\varepsilon_j \ge j$, hence $\varepsilon_j \to \infty$.[/step]

[guided]The role of the numbers $\varepsilon_j$ is to make the $j$th term visible only when $h$ is very small. We first specify the topology in which convergence must be proved. For a smooth family \begin{align*} b: T^*\mathbb{R}^n \times (0,h_0] \to \mathbb{C}, \end{align*} the symbol seminorms are \begin{align*} q_{\alpha,\beta}(b) := \sup_{(x,\xi,h) \in T^*\mathbb{R}^n \times (0,h_0]} \frac{|\partial_x^\alpha \partial_\xi^\beta b(x,\xi;h)|}{m(x,\xi)}. \end{align*} We enumerate these seminorms as $(q_k)_{k \ge 1}$. Controlling all $q_k$ is exactly controlling the $S(m)$ topology. Choose \begin{align*} \chi: \mathbb{R} \to \mathbb{R} \end{align*} with $\chi \in C_c^\infty(\mathbb{R})$, $\chi(t)=1$ for $|t| \le 1$, and $\chi(t)=0$ for $|t| \ge 2$. Because $h>0$, the condition $\chi(\varepsilon_j h)\ne 0$ implies $0 < h \le 2/\varepsilon_j$. Thus making $\varepsilon_j$ large forces the active values of $h$ to be small. Fix $j \ge 1$. We want the $j$th summand to be negligible after division by $h^N$ for every earlier order $N<j$, at least in the first $j$ seminorms. For $0 \le N < j$ and $1 \le k \le j$, no derivative in $q_k$ falls on $\chi(\varepsilon_j h)$ because $q_k$ differentiates only in $x$ and $\xi$. Hence \begin{align*} q_k\bigl(h^{-N} h^j \chi(\varepsilon_j h)a_j\bigr) \le \|\chi\|_\infty q_k(a_j)\sup_{0<h\le 2/\varepsilon_j} h^{j-N}. \end{align*} Since $j-N>0$, the supremum is \begin{align*} \sup_{0<h\le 2/\varepsilon_j} h^{j-N} = \left(\frac{2}{\varepsilon_j}\right)^{j-N}. \end{align*} Therefore \begin{align*} q_k\bigl(h^{-N} h^j \chi(\varepsilon_j h)a_j\bigr) \le \|\chi\|_\infty q_k(a_j)\left(\frac{2}{\varepsilon_j}\right)^{j-N}. \end{align*} For this fixed $j$, there are only finitely many pairs $(N,k)$ with $0 \le N < j$ and $1 \le k \le j$. We can therefore choose $\varepsilon_j$ so large that every one of these finitely many quantities is at most $2^{-j}$. Enlarging $\varepsilon_j$ again to ensure $\varepsilon_j \ge j$ preserves the inequalities and gives $\varepsilon_j \to \infty$.[/guided]

[step:Define the cutoff sum and prove that it is a symbol] Define \begin{align*} a: T^*\mathbb{R}^n \times (0,h_0] \to \mathbb{C} \end{align*} by \begin{align*} a(x,\xi;h) := \sum_{j=0}^{\infty} h^j \chi(\varepsilon_j h)a_j(x,\xi). \end{align*} For each fixed $h \in (0,h_0]$, only finitely many terms are nonzero: indeed, $\chi(\varepsilon_j h)\ne 0$ implies $\varepsilon_j \le 2/h$, and since $\varepsilon_j \to \infty$, this can hold for only finitely many $j$. Hence $a$ is a well-defined smooth family in $(x,\xi)$. We prove $a \in S(m)$. Fix $k \ge 1$. Split \begin{align*} a = \sum_{j=0}^{k} h^j \chi(\varepsilon_j h)a_j + \sum_{j=k+1}^{\infty} h^j \chi(\varepsilon_j h)a_j. \end{align*} The first sum is finite, so its $q_k$ seminorm is finite. For the tail, use the defining estimate with $N=0$. Since $j>k$ implies $k \le j$, we have \begin{align*} q_k\bigl(h^j \chi(\varepsilon_j h)a_j\bigr) \le 2^{-j}. \end{align*} Therefore \begin{align*} q_k\left(\sum_{j=k+1}^{\infty} h^j \chi(\varepsilon_j h)a_j\right) \le \sum_{j=k+1}^{\infty} 2^{-j} < \infty. \end{align*} Thus $q_k(a)<\infty$ for every $k$, so $a \in S(m)$. [/step]

[step:Control the tail after subtracting a finite partial sum] Fix $N \in \mathbb{N}_0$. Define \begin{align*} r_N: T^*\mathbb{R}^n \times (0,h_0] \to \mathbb{C} \end{align*} by \begin{align*} r_N(x,\xi;h) := h^{-N}\sum_{j=N}^{\infty} h^j \chi(\varepsilon_j h)a_j(x,\xi). \end{align*} We show that $r_N \in S(m)$. Fix $k \ge 1$. The term $j=N$ has finite $q_k$ seminorm because it equals $\chi(\varepsilon_N h)a_N$ when $N=j$, and $a_N \in S(m)$. The finitely many terms with $N<j<k$ also have finite $q_k$ seminorms. For the remaining terms, $j \ge \max\{k,N+1\}$ implies $k \le j$ and $N<j$, so the choice of $\varepsilon_j$ gives \begin{align*} q_k\bigl(h^{-N}h^j\chi(\varepsilon_jh)a_j\bigr) \le 2^{-j}. \end{align*} Consequently, \begin{align*} q_k(r_N) < \infty. \end{align*} Since $k$ was arbitrary, $r_N \in S(m)$. [/step]

[step:Show that the low-order cutoff errors are also of order $h^N$] For the same fixed $N$, define \begin{align*} e_N: T^*\mathbb{R}^n \times (0,h_0] \to \mathbb{C} \end{align*} by \begin{align*} e_N(x,\xi;h) := h^{-N}\sum_{j=0}^{N-1} h^j\bigl(\chi(\varepsilon_j h)-1\bigr)a_j(x,\xi). \end{align*} If $N=0$, this is the empty sum and hence $e_0=0$. Assume $N\ge 1$. Since $\chi(t)=1$ for $|t|\le 1$, the factor $\chi(\varepsilon_j h)-1$ can be nonzero only when $h \ge 1/\varepsilon_j$. Hence, for $0 \le j \le N-1$ and every seminorm $q_k$, \begin{align*} q_k\bigl(h^{j-N}(\chi(\varepsilon_j h)-1)a_j\bigr) \le (1+\|\chi\|_\infty)q_k(a_j)\varepsilon_j^{N-j}. \end{align*} There are only finitely many indices $0 \le j \le N-1$, so $q_k(e_N)<\infty$ for every $k$. Therefore $e_N \in S(m)$. [/step]

[step:Combine the tail and cutoff errors to obtain the asymptotic expansion] Using the definition of $a$, we compute \begin{align*} a - \sum_{j=0}^{N-1}h^j a_j = \sum_{j=N}^{\infty} h^j\chi(\varepsilon_jh)a_j + \sum_{j=0}^{N-1}h^j\bigl(\chi(\varepsilon_jh)-1\bigr)a_j. \end{align*} By the previous two steps, the first sum equals $h^N r_N$ with $r_N \in S(m)$, and the second sum equals $h^N e_N$ with $e_N \in S(m)$. Hence \begin{align*} a - \sum_{j=0}^{N-1}h^j a_j = h^N(r_N+e_N) \in h^N S(m). \end{align*} This proves the required asymptotic expansion for every $N \in \mathbb{N}_0$. [/step]

[step:Prove uniqueness modulo $h^\infty S(m)$] Let \begin{align*} a,b: T^*\mathbb{R}^n \times (0,h_0] \to \mathbb{C} \end{align*} be two symbols in $S(m)$ satisfying the same asymptotic conditions with coefficients $(a_j)_{j\ge0}$. Fix $N \in \mathbb{N}_0$. By hypothesis, \begin{align*} a-\sum_{j=0}^{N-1}h^j a_j \in h^N S(m) \end{align*} and \begin{align*} b-\sum_{j=0}^{N-1}h^j a_j \in h^N S(m). \end{align*} Subtracting these two inclusions gives \begin{align*} a-b \in h^N S(m). \end{align*} Since this holds for every $N \in \mathbb{N}_0$, \begin{align*} a-b \in \bigcap_{N=0}^{\infty} h^N S(m)=h^\infty S(m). \end{align*} Thus the asymptotic sum is unique modulo $h^\infty S(m)$. [/step]