I have a discrete times birth and death process $\{\Psi_n\}_{n\in \mathbb N}$ with birth probability $p$ and death probability $q$ defined as follows: \begin{align} \Psi_n=\sum_{i=1}^n\eta_i \end{align} where $\{\eta_i\}_{i\in\mathbb N}$ is a family of random variables i.i.d. such that \begin{align} \eta_i= \begin{cases} 1,\; &p\\ -1, \; &q\\ \end{cases} \end{align} and such that $\Psi_0=1$. Call \begin{align} V_N=\min\{n\geq 0: \Psi_n\geq N^{\frac{1}{10}}\} \end{align}
I need to prove that \begin{align}\label{p} \lim_{N\to +\infty}\mathbb P({V_{N}\leq N^{\frac{2}{10}}})=1 \end{align}
My idea is to apply the law of large numbers. From the law of large numbers I know that \begin{align} \frac{\Psi_{N^{\frac{2}{10}}}}{(p-q)N^{\frac{2}{10}}}\xrightarrow[N\to+\infty]{}1 \end{align} almost surely. Then, this allows me to conclude that \begin{align} \frac{\Psi_{N^{\frac{2}{10}}}}{N^{\frac{1}{10}}}\xrightarrow[N\to+\infty]{}+\infty \end{align} and consequently
$$\mathbb P(\Psi_{N^{\frac{2}{10}}}\geq N^{1/10})\xrightarrow[N\to +\infty]{} 1$$. This finally implies that \begin{align} \lim_{N\to +\infty}\mathbb P({V_{N}\leq N^{\frac{2}{10}}})=1 \end{align}
Is that correct in your opinion?