= Solution
Put $\epsilon=l/R$ and $a=1-\cos\theta_0$. Then
$$
\frac{r_B}{R}=\left(1+2a\epsilon+2a\epsilon^2\right)^{1/2}.
$$
The <binomial series> gives
$$
\frac{R}{r_B}=1-a\epsilon+\left(\frac32a^2-a\right)\epsilon^2+O(\epsilon^3),
$$
so
$$
v_2^2=2gR\left[a\epsilon+\left(a-\frac32a^2\right)\epsilon^2+O(\epsilon^3)\right].
$$
Since $v_1^2=2gRa\epsilon$, a second <binomial series> expansion yields
$$
\frac{v_2}{v_1}=1+\frac12\left(1-\frac32a\right)\epsilon+O(\epsilon^2)
=1+\left(-\frac14+\frac34\cos\theta_0\right)\frac lR+O\!\left(\frac{l^2}{R^2}\right).
$$
Thus the constants in the stated <Big O notation> expansion are
$$
\boxed{A=-\frac14,\qquad B=\frac34}.
$$
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