\\
                                          \end{pmatrix}y_T^{2k-1-i}\cdot(1+o_k(1))~,\end{aligned}$$ where this $o_k(1)$ now (depending on $k$) tends to zero when $k\rightarrow \infty$. For a moment, suppose we have already got , then we are able to control the largest eigenvalue of $A_T$, see the following theorem.

\[cb\] Under the same conditions as in Theorem \[th1\], the largest eigenvalue of $A_T$ converges to the right endpoint $b$ defined in almost surely.

(of Theorem \[cb\]) First we show that almost surely, $$\begin{aligned}
\label{ll1}
\liminf l_1 \ge b~.\end{aligned}$$ Indeed on the set $\{\liminf l_1<b\}$, we have $\liminf l_1<b-\delta$ for some $\delta=\delta(\omega)>0$. Let $g:\mathbb{R}\rightarrow \mathbb{R}_{+}$ be a continuous and positive function supported on $[b-\delta,b]$, with $\int g(x)dF(x)=1$, where $F$ is the LSD of $F^{A_T}$. Then $$\begin{aligned}
\liminf \int g(x)dF^{A_T}(x) \le 0~.\end{aligned}$$ For such $\omega$, $F^{A_T}$ will not converge weakly to $F$. Since this convergence occurs almost surely by Theorem \[th1\], the Claim is proved.

Next, we claim that for any $\delta>0$, $$\begin{aligned}
\label{ll2}
\sum_{p=1}^\infty P(l_1>b+\delta)<\infty~.\end{aligned}$$ Write $$l_1-b=(l_1-b)^{+}-(l_1-b)^{-}~,$$ with its positive and negative parts. Claim implies that almost surely, $(l_1-b)^{+} \to 0$. Therefore, a.s. $$\limsup (l_1-b) \le \limsup (l_1-b)^{+}=0~.$$ Combine with , we have a.s. $\lim (l_1-b)=0$. It remains to prove Claim .

Since we have $$\begin{aligned}
\label{l}
&\quad ~P(l_1>b+\delta)\leq P({\mathop{\text{tr}}}A_T^k\geq (b+\delta)^k)\leq\frac{{\mathbb{E}}{\mathop{\text{tr}}}A_T^k}{(b+\delta)^k}
