If $K$ is large enough, then for $n$ large enough, $\lambda_h^{Kn}/n^3 > (\lambda_h - \varepsilon)^{(K+\ell)(n+1)}$, so that

$$ \max_{a \ge h} \mathrm{P}_a^{\mathrm{T}_d} \left( |\mathcal{Z}_{(K+\ell)n}^h| < (\lambda_h - \varepsilon)^{(K+\ell)(n+1)} |\mathcal{Z}_{\ell n}^h \ne \emptyset \right) \le c e^{-c'n}. $$

We adjust the conditioning: $\{\mathcal{Z}_{(K+\ell)n}^h \ne \emptyset\} \subset \{\mathcal{Z}_{\ell n}^h \ne \emptyset\}$, and $|\mathcal{Z}_{(K+\ell)n}^h| < (\lambda_h - \varepsilon)^{(K+\ell)(n+1)}$ on $\{\mathcal{Z}_{\ell n}^h \ne \emptyset\} \setminus \{\mathcal{Z}_{(K+\ell)n}^h \ne \emptyset\}$. Therefore, there exists $n_0 \ge 1$ such that for all $n \ge n_0$,

$$ \max_{a \ge h} \mathrm{P}_a^{\mathrm{T}_d} \left( |\mathcal{Z}_{(K+\ell)n}^h| < (\lambda_h - \varepsilon)^{(K+\ell)(n+1)} |\mathcal{Z}_{(K+\ell)n}^h \ne \emptyset \right) \le c \exp(-c'(K+\ell)(n+1)), \quad (A.12) $$

where the new value of $c'$ depends on the constants $K$ and $\ell$. This yields (A.9) for large enough multiples of $K + \ell$. One can readily replace each $\mathcal{Z}_m^h$ by $\mathcal{Z}_m^{h,+}$ in this reasoning (for any $m \ge 1$), to get the same result for $\mathcal{Z}_{(K+\ell)n}^{h,+}$ instead of $\mathcal{Z}_{(K+\ell)n}^h$.

It remains to show the result for non multiples of $K + \ell$. Let $k \ge (K + \ell)n_0$. Write $k = (K + \ell)n + m$, with $0 \le m \le (K + \ell) - 1$. Note that on $\{\mathcal{Z}_k^h \ne \emptyset\}$, $\mathcal{Z}_m^h$ has at least one vertex whose $(k-m)$-offspring is not empty. Hence

$$ \begin{aligned} \max_{a \ge h} \mathrm{P}_a^{\mathrm{T}_d} (|\mathcal{Z}_k^h| < (\lambda_h - \varepsilon)^k |\mathcal{Z}_k^h \ne \emptyset) & \le \max_{a \ge h} \mathrm{P}_a^{\mathrm{T}_d} \left( |\mathcal{Z}_{k-m}^{h,+}| < (\lambda_h - \varepsilon)^k |\mathcal{Z}_{k-m}^{h,+} \ne \emptyset \right) \\ & \le \max_{a \ge h} \mathrm{P}_a^{\mathrm{T}_d} \left( |\mathcal{Z}_{k-m}^{h,+}| < (\lambda_h - \varepsilon)^{(k-m)+(K+\ell)} |\mathcal{Z}_{k-m}^{h,+} \ne \emptyset \right) \\ & \le c \exp(-c'(K+\ell)(n+1)) \\ & \le c e^{-c'k}, \end{aligned} $$

where the third inequality comes from (A.12). Adapting this last computation for $\mathcal{Z}_k^{h,+}$ is immediate. This concludes the proof of (3.2). $\square$

### A.3 Proof of Proposition 4.1

In the proof, we will use the following two facts on tree excesses, that we use implicitly in other parts of the paper. First, for any subgraph $A$ of any finite graph $G$ and $R \in \mathbb{N}$, $\mathrm{tx}(B_G(A, R)) \ge \mathrm{tx}(A)$, with equality if and only if $B_G(A, R)$ has the same number of cycles and the same number of connected components as $A$. Second, if $G$ is connected, $\mathrm{tx}(G) = 0$ if and only if $G$ is a tree, i.e. has no cycle.

Let us give a short proof. For the first fact, it is enough to show that for any choice of $A$ and $G$, the sequence $(\mathrm{tx}(B_G(A, R)))_{R \ge 0}$ is non-decreasing. Since $B_G(A, R+1) = B_G(B_G(A, R), 1)$, it is enough to check that $\mathrm{tx}(B_G(A, 1)) \ge \mathrm{tx}(A)$, without loss of generality. Let $a_1, \dots, a_n$ be the vertices of $A$ for some $n \ge 1$. For $i = 1, \dots, n$, add to $A$ the neighbours of $a_i$ and the corresponding edges. This yields $B_G(A, 1)$. Each time we add a neighbour $x$ of $a_i$, if $x$ was not already in $A$, then the edge $\{a_i, x\}$ was not either. Hence, the quantity (number of edges - number of vertices) is not decreasing during this process, so that $\mathrm{tx}(B_G(A, 1)) \ge \mathrm{tx}(A)$. As for the second fact, it is classical that if $G$ is a tree, then $\mathrm{tx}(G) = 0$ (see e.g. Theorem 2.2 in [9]). Reciprocally, if $G$ is connected, is not a tree and $\mathrm{tx}(G) = 0$, then there is a spanning tree $T$ of $G$ such that $\mathrm{tx}(T) < 0$ (since $T$ is obtained from $G$ by deleting no vertex and at least one edge on a cycle of $G$), but this contradicts the fact that if $T$ is a tree, then $\mathrm{tx}(T) = 0$.

We will also need the lemma below, which is a consequence of Lemma 3.3 in [12] and of the following observation. If $(X_j)_{j \ge 0}$ is a SRW on $\mathrm{T}_d$, then the trajectory of its height $(\mathbb{h}_{\mathrm{T}_d}(X_j))_{j \ge 0}$ is distributed as a random walk on $\mathbb{N} \cup \{0\}$ with transition probabilities $1/d$ towards the left neighbour and $(d-1)/d$ towards the right neighbour, and reflected at 0.

**Lemma A.4 (Geometric repulsion).** Let $s \in \mathbb{N}$, and $A \subset V_n$ such that $\mathrm{tx}(B_{\mathcal{M}_n}(A, s)) = \mathrm{tx}(A)$. Let $x \in V_n \setminus B_{\mathcal{M}_n}(A, s)$, let $(X_j)_{j \ge 0}$ be a SRW started at $x$ and $\tau$ its first hitting
time of A. Then $\tau$ dominates stochastically a geometric random variable of parameter $(d-1)^{-s}$.

We give a shortened proof of Proposition 4.1, due to the many similarities with the proof of Proposition 2.7 in [4], and refer the reader to [4] for details.

**Proof of Proposition 4.1.** Let us first prove (4.2). By our assumptions on A and $y T_y := B_{M_n}(y, \bar{y}, a_n - 1)$ is a tree rooted at $y$. Let $\partial T_y$ be the $(a_n - 1)$-offspring of $y$ in $T_y$. Let $\tau$ (resp. $\tau'$) be the hitting time of $\partial T_y$ (resp. $\partial T_y \cup \{y\}$) by a SRW $(X_j)_{j \ge 0}$. By Proposition 2.3,

$$ \mathbb{E}^{M_n}[\psi_{M_n}(y)|\sigma(A)] = \mathbf{E}_y^{M_n}[\psi_{M_n}(X_{H_A}) \mathbf{1}_{H_A \le \tau}] + \mathbf{E}_y^{M_n}[\psi_{M_n}(X_{H_A}) \mathbf{1}_{H_A > \tau}] \\ - \frac{\mathbf{E}_y^{M_n}[H_A]}{\mathbf{E}_{\pi_n}^{M_n}[H_A]} \mathbf{E}_{\pi_n}[\psi_{M_n}(X_{H_A})]. \quad (A.13) $$

By the reasoning leading to the domination of $U_{A,x}^{G_n}$ below (2.43) in [4], we get

$$ \left| \mathbf{E}_y^{M_n} [\psi_{M_n}(X_{H_A}) \mathbf{1}_{H_A \le \tau}] - \frac{1}{d-1} \psi_{M_n}(\bar{y}) \right| \le \log^{-4} n. $$

In particular, it is worth noting that in our context, (2.33) in [4] becomes

$$ \left| \mathbf{P}_y^{M_n}(H_A \le \tau) - \frac{1}{d-1} \right| \le \log^{-6} n \quad \text{and} \quad \mathbf{E}_y^{M_n}[\tau'] \le 2 \log n. \quad (A.14) $$

Therefore, to establish (4.2), it is enough to show that

$$ \left| \mathbf{E}_y^{M_n} [\psi_{M_n}(X_{H_A}) \mathbf{1}_{H_A > \tau}] - \frac{d-2}{d-1} \mathbf{E}_{\pi_n}^{M_n} [\psi_{M_n}(X_{H_A})] \right| \le 3 \log^{-5} n \quad (A.15) $$

and

$$ \left| \frac{d-2}{d-1} - \frac{\mathbf{E}_y^{M_n}[H_A]}{\mathbf{E}_{\pi_n}^{M_n}[H_A]} \right| \mathbf{E}_{\pi_n}^{M_n}[\psi_{M_n}(X_{H_A})] \le \log^{-5} n. \quad (A.16) $$

We start with (A.15). Following the proofs of (2.35) (using Lemma A.4 with $s = a_n$, fixing $\kappa$ large enough, instead of Lemma 3.4 of [12]) and (2.36) in [4], and using the fact that $\max_{z \in A} |\psi_{M_n}(z)| \le \log^{2/3} n$, we get that

$$ \sup_{z \in \partial T_y} \left| \mathbf{E}_z^{M_n} [\psi_{M_n}(X_{H_A})] - \mathbf{E}_{\pi_n}^{M_n} [\psi_{M_n}(X_{H_A})] \right| \le \log^{-6} n. $$

Also, by (A.14), $\left| \mathbf{P}_y^{M_n}(H_A > \tau) - \frac{d-2}{d-1} \right| \le \log^{-6} n$. Combining these two facts yields (A.15).

As for (A.16), we have $\mathbf{E}_y^{M_n}[H_A] = \mathbf{E}_y^{M_n}[\tau'] + \sum_{z \in \partial T_y} \mathbf{P}_y^{M_n}(X_{\tau'} = z) \mathbf{E}_z^{M_n}[H_A]$. By (3.20) of [12], $\mathbf{E}_{\pi_n}^{M_n}[H_A] \ge \frac{n}{4|A|} \ge \log^8 n/4$. Combining this with (A.14), we get

$$ \left| \frac{\mathbf{E}_y^{M_n}[H_A]}{\mathbf{E}_{\pi_n}^{M_n}[H_A]} - \sum_{z \in \partial T_y} p'_z \frac{\mathbf{E}_z^{M_n}[H_A]}{\mathbf{E}_{\pi_n}^{M_n}[H_A]} \right| \le \frac{8 \log n}{\log^8 n} \le \log^{-6} n, $$

where $p'_z := \mathbf{P}_y^{M_n}(X_{\tau'} = z)$. By (A.14), $\left| \sum_{z \in \partial T_y} p'_z - \frac{d-2}{d-1} \right| \le \log^{-6} n$. Therefore,

$$ \left| \frac{\mathbf{E}_y^{M_n}[H_A]}{\mathbf{E}_{\pi_n}^{M_n}[H_A]} - \frac{d-2}{d-1} \right| \le 2 \log^{-6} n + \max_{z \in \partial T_y} \left| \frac{\mathbf{E}_z^{M_n}[H_A]}{\mathbf{E}_{\pi_n}^{M_n}[H_A]} - 1 \right|. $$

Since $\max_{z \in A} |\psi_{M_n}(z)| \le \log^{2/3} n$, (A.16) and thus (4.2) follows once we show that

$$ \max_{z \in \partial T_y} \left| \frac{\mathbf{E}_z^{M_n}[H_A]}{\mathbf{E}_{\pi_n}^{M_n}[H_A]} - 1 \right| \le 5 \log^{-6} n. \quad (A.17) $$
Here lies the main difference with Proposition 2.7 in [4], which makes the stronger assumption that $|A| = O(\log n)$. For all $z \in \partial T_y$, following the first part of the proof of Proposition 3.5 in [12], we obtain the upper bound:

$$ 
\begin{aligned}
\mathbf{E}_{z}^{\mathcal{M}_n}[H_A] & \le \log^2 n + \sum_{z' \in V_n} (\pi_n(z') + e^{-\lambda_{\mathcal{M}_n} \log^2 n}) \mathbf{E}_{z'}^{\mathcal{M}_n}[H_A] \\
& \le \log^2 n + (1 + n e^{-\lambda_{\mathcal{M}_n} \log^2 n}) \mathbf{E}_{\pi_n}^{\mathcal{M}_n}[H_A].
\end{aligned}
\quad (A.18) $$

Recalling that $\mathbf{E}_{\pi_n}^{\mathcal{M}_n}[H_A] \ge \log^8 n/4$; this yields

$$ \frac{\mathbf{E}_{z}^{\mathcal{M}_n}[H_A]}{\mathbf{E}_{\pi_n}^{\mathcal{M}_n}[H_A]} \le \frac{4 \log^2 n}{\log^8 n} + 1 + n e^{-\lambda_{\mathcal{M}_n} \log^2 n} \le 1 + 5 \log^{-6} n. \quad (A.19) $$

Conversely, the second part of the proof of [12] gives us that

$$ \frac{\mathbf{E}_{z}^{\mathcal{M}_n}[H_A]}{\mathbf{E}_{\pi_n}^{\mathcal{M}_n}[H_A]} \ge 1 - n e^{-\lambda_{\mathcal{M}_n} \log^2 n} - \mathbf{P}_z^{\mathcal{M}_n}(H_A \le \log^2 n)(1 + 5 \log^{-6} n). $$

By Lemma A.4 applied as below (A.16), we have $\sup_{z \in \partial T_y} \mathbf{P}_z^{\mathcal{M}_n}(H_A \le \log^2 n) \le \log^{-6} n$. Together with (I) and (A.19), this yields (A.17) and the proof of (4.2) is complete (note that the required lower bounds on $\kappa$ given by Lemma A.4 are uniform in $y$ and $A$).

We prove (4.3) in the same fashion. Since $H_A \ge \tau'$ a.s., (2.11) yields

$$ \left| \mathrm{Var}^{\mathcal{M}_n}(\psi_{\mathcal{M}_n}(y) | \sigma(A)) - \frac{d}{d-1} \right| \le \left| G_{\mathcal{M}_n}(y, y) - \mathbf{E}_y^{\mathcal{M}_n} [G_{\mathcal{M}_n}(y, X_{H_A}) \mathbf{1}_{H_A=\tau'}] - \frac{d}{d-1} \right| \\ + \left| \mathbf{E}_y^{\mathcal{M}_n} [G_{\mathcal{M}_n}(y, X_{H_A}) \mathbf{1}_{H_A>\tau'}] - \frac{\mathbf{E}_y^{\mathcal{M}_n}[H_A]}{\mathbf{E}_{\pi_n}[H_A]} \mathbf{E}_{\pi_n}^{\mathcal{M}_n}[G_{\mathcal{M}_n}(y, X_{H_A})] \right|. \quad (A.20) $$

Following the reasoning at (2.47) and below in Proposition 2.7 of [4], one shows that the first term of the RHS is $O(\log^{-5} n)$. Then, we apply to the second term of the RHS of (A.20) the same reasoning as (A.15) and (A.16), the inequality $\max_{x,y \in V_n} |G_{\mathcal{M}_n}(x, y)| \le K_5$ (by (2.4), since $\mathcal{M}_n$ is a good graph) replacing the inequality $\max_{z \in A} |\psi_{\mathcal{M}_n}(z)| \le \log^{2/3} n$. $\square$

## A.4 Proof of Lemmas 7.6 and 7.7

*Proof of Lemma 7.6.* We follow the argument of Proposition 4.1, with a few adjustments.

First, the bounds in the first inequality of (A.14) and in (A.15) are in fact $e^{-ca_n}$ for some constant $c > 0$, and one can replace $a_n$ by $r_n$.

Second, the proof of (A.16) can be adapted by noting the following two facts. On one hand, the condition $|A| \le n^{2/3}$ implies that $\mathbf{E}_{\pi_n}[H_A] \ge n^{1/3}/4$. On the other hand, the second inequality of (A.14) still holds for large enough $n$, using the same comparison with a biased random walk as in (2.33) in [4], since for $k \ge 0.3 \log n$ and a constant $\gamma > 0$ that depends neither on $k$ nor on $n$, $P(\tau' \ge k) \le e^{-\gamma k}$. $\square$

*Proof of Lemma 7.7.* We will only show (7.32) and (7.33). The other proofs are very similar and left to the reader.

We start with the proof of (7.32). We follow the proof scheme of (4.2). Let $\tau$ be the hitting time of $\partial B_{\mathcal{M}_n}(y, \bar{y}, r_n) \setminus \{y\}$ by a SRW $(X_k)_{k \ge 0}$. Note that $\{H_A \le \tau\} \subseteq \{X_{H_A} = y\}$. We write

$$ \begin{aligned} \mathbf{E}^{\mathcal{M}_n}[\psi_{\mathcal{M}_n}(y) | \sigma(A)] = & \mathbf{P}_y^{\mathcal{M}_n}(X_{H_A} = \bar{y}, H_A \le \tau) \psi_{\mathcal{M}_n}(\bar{y}) + \mathbf{E}_y^{\mathcal{M}_n}[\psi_{\mathcal{M}_n}(X_{H_A}) \mathbf{1}_{H_A > \tau}] \\ & - \frac{\mathbf{E}_y^{\mathcal{M}_n}[H_A]}{\mathbf{E}_{\pi_n}^{\mathcal{M}_n}[H_A]} \mathbf{E}_{\pi_n}^{\mathcal{M}_n}[\psi_{\mathcal{M}_n}(X_{H_A})]. \end{aligned} $$
Since $B_{M_n}(y, \bar{y}, r_n)$ and $B_{G_m}(z_k, \bar{z}_k, r_n)$ are isomorphic,

$$ \mathbf{P}_{y}^{M_n}(X_{H_A} = \bar{y}, H_A \le \tau) = \mathbf{P}_{z_k}^{G_m}(X_{H_{\partial B_{G_m}(z_k, \bar{z}_k, r_n)}} = \bar{z}_k) := \alpha_k''. $$

Since

$$ \{X_{H_{\partial B_{G_m}(z_k, \bar{z}_k, r_n)}} = \bar{z}_k\} \subseteq \{H_{\{\bar{z}_k\}} < +\infty\}, \text{ we have } \alpha_k'' \le \alpha_k'. $$

Reciprocally, on $\{H_{\{\bar{z}_k\}} < +\infty\} \setminus \{X_{H_{\partial B_{G_m}(z_k, \bar{z}_k, r_n)}} = \bar{z}_k\}$, a SRW starting at distance $r_n$ of $C_m$ has to reach $C_m$. As in the proof of (A.14), a comparison with a biased SRW on $Z$ shows that this happens with a probability $O(e^{-cr_n})$ for some constant $c > 0$ uniquely depending on $d$ and we get that if a small enough, then for large enough $n$, $|\mathbf{P}_{y}^{M_n}(X_{H_A} = \bar{y}, H_A \le \tau) - \alpha_k'| \le n^{-3a}$.

It remains to establish

$$ \left| \mathbf{E}_{y}^{M_n} [\psi_{M_n}(X_{H_A}) \mathbf{1}_{H_A > \tau}] - \frac{\mathbf{E}_{y}^{M_n}[H_A]}{\mathbf{E}_{\pi_n}^{M_n}[H_A]} \mathbf{E}_{\pi_n}^{M_n}[\psi_{M_n}(X_{H_A})] \right| \le n^{-3a}. \quad (A.21) $$

To do so, one adapts the proofs of (A.15) and (A.16) exactly as in the proof of Lemma 7.6: if $H_A > \tau$, $(X_k)_{k \ge 0}$ leaves $B_{M_n}(A \cup C \cup P, r_n)$ before hitting $A$. By Lemma A.4 with $s = r_n$, if a is small enough, $(X_k)$ does not hit $A$ within the next $\log^2 n$ steps with probability at least

$$ (1 - (d - 1)^{-r_n})^{\log^2 n} \ge 1 - 2 \log^2 n (d - 1)^{-r_n} \ge 1 - n^{-4a}. $$

Then, we use Corollary 2.1.5 of [29] as in the proof of (2.36) of [4]: after $\lfloor \log^2 n \rfloor$ steps, the fact that $M_n$ is an expander (i.e. $\lambda_{M_n} > K_3 > 0$) forces the empirical distribution of $X_k$ to be very close to the uniform distribution $\pi_n$. (7.32) follows.

For (7.33), we follow the proof scheme of (4.3). If $\tau'$ is the exit time of $B_{M_n}(y, \bar{y}, r_n)$, we have

$$ \begin{aligned} |\mathrm{Var}^{M_n}(\psi_{M_n}(y)|\sigma(A)) - \gamma_k'| & \le |G_{M_n}(y, y) - \mathbf{E}_y^{M_n}[G_{M_n}(y, X_{H_A}) \mathbf{1}_{H_A=\tau'}] - \gamma_k'| \\ & \quad + \left| \mathbf{E}_y^{M_n}[G_{M_n}(y, X_{H_A}) \mathbf{1}_{H_A>\tau'}] - \frac{\mathbf{E}_y^{M_n}[H_A]}{\mathbf{E}_{\pi_n}[H_A]} \mathbf{E}_{\pi_n}^{M_n}[G_{M_n}(y, X_{H_A})] \right|. \end{aligned} \quad (A.22) $$

We deal with the second term of the RHS as (A.21) to show that it is $O(n^{-3a})$. As for the first term, we have as in (2.47) and below in [4]:

$$ \mathbf{E}_{y}^{M_n}[G_{M_n}(y, X_{H_A}) \mathbf{1}_{H_A=\tau'}] = \mathbf{E}_{y}^{M_n}[G_{M_n}(y, X_{\tau'})] - O(n^{-3a}) $$

if a is small enough, by (2.4). Now, by (A.2) applied to $D := B_{M_n}(y, \bar{y}, r_n)$ (note that $T_D = \tau'$) and the second inequality of (A.14) (which still holds, as remarked in the proof of Lemma 7.6), we get

$$ |G_{M_n}(y, y) - \mathbf{E}_y^{M_n}[G_{M_n}(y, X_{\tau'})] - G_{M_n}^D(y, y)| \le \frac{\mathbf{E}_y^{M_n}[\tau']}{n} = O(n^{-3a}). $$

But $D$ is isomorphic to $E := B_{G_m}(z_k, \bar{z}_k, r_n)$, so that

$$ \begin{aligned} G_{M_n}^D(y, y) &= G_{G_m}^E(z_k, z_k) \\ &= G_{G_m}(z_k, z_k) - \mathbf{P}_{z_k}^{G_m}(T_E = \bar{z}_k) G_{G_m}(z_k, \bar{z}_k) - \mathbf{P}_{z_k}^{G_m}(T_B \ne \bar{z}_k) G_{G_m}(z_k, z) \end{aligned} $$

for any $z \in \partial B_{G_m}(E, 1) \setminus \{\bar{z}_k\}$, by cylindrical symmetry of $B_{G_m}(E, 1)$. One checks easily that if a is small enough, then for $n$ large enough,

$$ G_{M_n}^D(y, y) = G_{G_m}^E(z_k, z_k) = G_{G_m}(z_k, z_k) - \mathbf{P}_{z_k}^{G_m}(T_E = \bar{z}_k) G_{G_m}(z_k, \bar{z}_k) + O(n^{-3a}). $$
One easily adapts the reasoning leading to (A.14), despite the presence of one cycle, to get

$$ \mathbf{P}_{z_k}^{G_m}(T_E = \bar{z}_k) = \mathbf{P}_{z_k}^{G_m}(H_{\{\bar{z}_k\}} < +\infty) + O(n^{-3a}) \text{ for a small enough.} $$

Note indeed that $\{T_E = \bar{z}_k\} \subseteq \{H_{\{\bar{z}_k\}} < +\infty\}$. Reciprocally, if $z \in \partial B_{G_m}(E, 1) \setminus \{\bar{z}_k\}$, a SRW starting at $z$ has a probability decaying exponentially with $r_n$ to reach $\bar{z}_k$, since there are at most two injective paths from $z$ to $\bar{z}_k$, and each contains at least $r_n - 3$ vertices where the SRW has a positive probability (only depending on $d$) to enter a subtree isomorphic $T_d^+$ and to never leave it.

Since $\gamma'_k = G_{G_m}(z_k, z_k) - \mathbf{P}_{z_k}^{G_m}(H_{\{\bar{z}_k\}} < +\infty)G_{G_m}(z_k, \bar{z}_k)$, we obtain

$$ |G_{M_n}^D(y, y) - \gamma'_k| = O(n^{-3a}). $$

All in all, we get that the first term of the RHS of (A.22) is $O(n^{-3a})$, and (7.33) follows. □

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