Probability with Martingales:
I tried writing out the details of the proof avoiding $\ge$ if I felt it was unnecessary. Please tell me if I got any steps wrong.
$$\liminf \frac{1}{b_n} \sum_{k=1}^{n} (b_k - b_{k-1})v_k$$
$$\color{red}{=} \liminf \frac{1}{b_n} \left[\sum_{k=1}^{N} (b_k - b_{k-1})v_k + \sum_{k=N+1}^{n} (b_k - b_{k-1})v_k \right] \tag{1}$$
$$\color{red}{>} \liminf \frac{1}{b_n} \left[\sum_{k=1}^{N} (b_k - b_{k-1})v_k + \sum_{k=N+1}^{n} (b_k - b_{k-1})(v_{\infty} - \epsilon) \right] \tag{2}$$
$$\color{red}{=} \liminf \frac{1}{b_n} \left[\sum_{k=1}^{N} (b_k - b_{k-1})v_k + (v_{\infty} - \epsilon)\sum_{k=N+1}^{n} (b_k - b_{k-1}) \right] \tag{3}$$
$$\color{red}{=} \liminf \frac{1}{b_n} \left[\sum_{k=1}^{N} (b_k - b_{k-1})v_k + (v_{\infty} - \epsilon)(b_n - b_{N+1}) \right] \tag{4}$$
$$\color{red}{\ge} \liminf \frac{1}{b_n} \left[\sum_{k=1}^{N} (b_k - b_{k-1})v_k\right] + \liminf \frac{1}{b_n} \left[(v_{\infty} - \epsilon)(b_n - b_{N+1}) \right] \tag{5}$$
$$\color{red}{=} 0 + \liminf \frac{1}{b_n} \left[(v_{\infty} - \epsilon)(b_n - b_{N+1}) \right] \tag{6}$$
$$\color{red}{=} 0 + (v_{\infty} - \epsilon) \tag{7}$$
On $(6)$:
$$\liminf \frac{1}{b_n} \left[\sum_{k=1}^{N} (b_k - b_{k-1})v_k\right] \stackrel{?}{=} \frac{1}{\liminf b_n} \left[\sum_{k=1}^{N} (b_k - b_{k-1})v_k\right]$$
On $(7)$:
$$\liminf \frac{1}{b_n} \left[(v_{\infty} - \epsilon)(b_n - b_{N+1}) \right]$$
$$ \color{red}{=} (v_{\infty} - \epsilon) \liminf \frac{1}{b_n} \left[b_n - b_{N+1} \right] \tag{7.1}$$
$$ \color{red}{=} (v_{\infty} - \epsilon) \liminf \left[1 - \frac{b_{N+1}}{b_n} \right] \tag{7.2}$$
$$ \color{red}{=} (v_{\infty} - \epsilon) \left[1 - \liminf \frac{b_{N+1}}{b_n} \right] \tag{7.3}$$
$$ \color{red}{=} (v_{\infty} - \epsilon) \left[1 - \frac{b_{N+1}}{\liminf b_n} \right] \tag{7.4}$$
In $(2)$, the sign $>$ should be replaced by $\ge$ (note that when taking the limit, $>$ will become $\ge$).
Notice that $\lim\inf\frac{C}{b_n} = 0$ as $b_n\to \infty$ for any constant $C$, it is straightforward to obtain $(6)$ and $(7)$ from $(5)$: \begin{align} &\liminf \frac{1}{b_n} \left[\sum_{k=1}^{N} (b_k - b_{k-1})v_k\right] + \liminf \frac{1}{b_n} \left[(v_{\infty} - \epsilon)(b_n - b_{N+1}) \right] \tag{5} \\ = &\liminf \frac{C_1}{b_n} + \liminf \left(v_{\infty} - \epsilon + \frac{- b_{N+1}(v_{\infty} - \epsilon)}{b_n} \right) \\ =& \liminf \frac{C_1}{b_n} + \liminf \left(\frac{C_2}{b_n} \right) + v_{\infty} - \epsilon \\ = & 0 + 0 + v_{\infty} - \epsilon \end{align} where $C_1 = \left[\sum_{k=1}^{N} (b_k - b_{k-1})v_k\right]$ and $C_2 = - b_{N+1}(v_{\infty} - \epsilon)$.