ProofLemma

\begin{proof}
The expressions defining $\theta(k), \theta_1(k), \theta_2(k), \theta_3(k), \theta_4$(k), and $\theta_5(k)$ have a similar behavior to infer the further inequalities (7a), (7b), (7c), (7d), (7e), and (7f); i.e., they can be written as $\frac{a_1(k) - \frac{a_1(k)}{a_2(k)}}{a_1(k) - a_3}$, with certain $a_1(k),a_2(k) \in O(k)$ and $a_3\in\mathbb{R}$ for each case. Let us choose $\theta(k)$ as example while proving (7a). First of all, we rewrite it as
\begin{equation}
    \theta(k) = \frac{\frac{k+m_1 - 1}{2} - \frac{k+m_1 -1}{2(k+m_2+l-1)}}{\frac{k+m_1 - 1}{2} + \frac1n - \frac12} = \frac{ 1 - \frac{1}{k+m_2+l-1}}{1 - \frac{n-2}{n(k+m_1-1)}}.
\end{equation}
Therefore, $\lim_{k\to\infty} \theta(k) = 1$ and, as a consequence, there exists $k_1 \in\mathbb{R}$ such that $\theta > 0$ for $k > k_1$. To prove that $\theta <1$ we notice that
\begin{equation}
    \frac{1}{k+m_2+l-1} > \frac{n-2}{n(k+m_1-1)} \Leftrightarrow \frac{k+m_1-1}{k+m_2+l-1} > \frac{n-2}{n} = 1 - \frac2n.
\end{equation}
This follows from
\begin{equation}
    \lim_{k\to\infty} \frac{k+m_1-1}{k+m_2+l-1} = 1 \Rightarrow \exists\ k_2 \in\mathbb{R}: \frac{k+m_1-1}{k+m_2+l-1} > 1 - \frac2n, \quad k>k_1.
\end{equation}
Therefore, $\theta \in (0,1)$ for all $k>\max\{k_1,k_2\}$.

As in the previous observation, $\sigma(k), \sigma_2(k), \sigma_3(k)$, and $\sigma_5(k)$ can be written as $\frac{a_2(k)}{a_1(k)}$ for each respective case. Therefore, (8a), (8b), (8c), and (8d) follow a similar reasoning; i.e., we can write them as $\frac{a_2(k) - 1}{a_1(k)-a_3}$ respectively. To prove (8a) we notice that:
\begin{equation}
    \frac{\sigma(k)\theta(k)}{2} = 1 - \frac{m_1 + \frac2n -m_2 -l}{k+m_1-2+\frac{2}{n}} \in (0,1) \Leftrightarrow \frac{m_1 + \frac2n -m_2 -l}{k+m_1-2+\frac{2}{n}} \in (0,1).
\end{equation}
Being $\lim_{k\to\infty} \frac{m_1 + \frac2n -m_2 -l}{k+m_1-2+\frac{2}{n}} = 0$, the inclusion holds for $m_1 + \frac2n > m_2 + l$ as claimed.

The last case leads, jointly with the hypothesis $\alpha \geq 1$, to the conclusion $\alpha < m_1 + \frac2n$ and consequently $m_1 > 1 - \frac2n$.
\end{proof}