=\frac{p\cdot{\mathbb{E}}(m_k(A_T))}{(b+\delta)^k}~,
$$ and by , $$\begin{aligned}
p\cdot{\mathbb{E}}(m_k(A_T))&=p\sum_{i=0}^{k-1}\frac 1k \begin{pmatrix}
                               2k \\
                               i \\
                             \end{pmatrix}\begin{pmatrix}
                                            k \\
                                            i+1 \\
                                          \end{pmatrix}y_T^{2k-1-i}\cdot(1+o_k(1))\\
&\leq \left(p^{1/k}b(y_T)\right)^k\cdot(1+o_k(1))\xrightarrow{p\rightarrow \infty} b^k~,\end{aligned}$$ where the last convergence is due to the choice of $k$ that $k=(\log p)^{1.01}$ (see Theorem \[largek\]), so the term $p^{1/k}\rightarrow 1$ when $p \rightarrow \infty$. Once we fixed this $\delta>0$ in , then the right hand side tends to $\left(\frac{b}{b+\delta}\right)^k$, which is summable.

The proof of Theorem \[cb\] is complete.

The remaining of the section is devoted to deriving used in the proof above.

\[largek\] Let the conditions in Theorem \[th1\] hold, and $k$ is an integer of size $k=(\log p)^{1.01}$ (say). Then we have $$\begin{aligned}
{\mathbb{E}}(m_k(A_T))&=\sum_{i=0}^{k-1}\frac 1k \begin{pmatrix}
                               2k \\
                               i \\
                             \end{pmatrix}\begin{pmatrix}
                                            k \\
                                            i+1 \\
                                          \end{pmatrix}y_T^{2k-1-i}\cdot(1+o_k(1))~.\end{aligned}$$

(of Theorem \[largek\]) When $k \rightarrow \infty$, the term $\#\{\text{isomorphism class}\}$ in is no more a constant order. Then the main task is to show that the contribution of Category 3 to the $k$-th moment of $A_T$ when $k\rightarrow \infty$ can still be negligible compared with Category 1.

Since in Category 3, we have $t+s-1<2k$. Using the previous notion of the characteristic sequence, we have $\#\{1\}=t-1$ and $\#\{-1\}=2k-s$, and this is equivalent to the fact that $\#\{1\}<\#\{-1\}$.

