Suppose we have a lazy random walk $(S_0, S_1, S_2,...)$ on $\mathbb{Z}$ starting from $S_0 =0$ biased to the right, i.e. $S_n = \sum_{k=0}^n X_k$ with $\mathbb{P}(X_1=1)=p$, $\mathbb{P}(X_1=0)=q$, $\mathbb{P}(X_1=-1)=r$ and $p>r$ and $p+q+r=1$. The $X_i$'s are also assumed to be i.i.d. with only exception $X_0=0$.
Can we evaluate the quantity $\mathbb{P}(T_{-1}>n)$ where $T_{-1} = \inf \{ k \geq 0: S_k=-1 \}$ is the first hitting time of $-1$ ? My guess is that $\mathbb{P}(T_{-1}=\infty)>0$, so I want to prove this fact (and obtain an expression for $\mathbb{P}(T_{-1}=\infty)$ depending on $p$ and $q$) by showing that $\lim_{n\to\infty} \mathbb{P}(T_{-1}>n) > 0$.
$\newcommand{\PM}{\mathbb{P}}$Your main interest is the expression for $\PM(T_{-1}=\infty)$? Then it is much easier to just go for that instead of calculating $\PM(T_{-1}>n)$ first. The reason for that is because $\PM(T_{-1}=\infty)$ is the probability of never hitting $-1$ starting at $0$ and it is easy to calculate the probability of ever hitting $-1$ starting at $0$. Define the following: \begin{align} f_n:=\PM(\text{ever hitting} -1 \text{ starting at } n) \end{align} Clearly $f_{-1}=1$ moreover one has $$\tag{1}f_0 = r + qf_0+ pf_1$$ Now I give you an exercise, namely to prove that $f_1 = f_0 f_0$. So $(1)$ becomes: $$f_0 = r + qf_0+ pf_0^2$$ This has two solutions, namely: \begin{align} f_0 = \frac{r}{p} \ \ \vee \ \ f_0 = 1 \end{align} Take $r=0$ and immediately see that $f_0=1$ fells off, hence $f_0 = \frac{r}{p}$. Finally: \begin{align} \PM(T_{-1}=\infty) = 1- f_0 = 1-\frac{r}{p} \end{align}