Infinite divergent series - does the convergence of quotient imply convergence of difference?

Your thoughts are correct. To form our intuition, consider the two functions $f(x)=x^2+2x$ and $g(x)=x^2$. Clearly $\lim_{x\to\infty}f(x)/g(x)=1$ while $f(x)-g(x)=2x\to\infty$ as $x\to\infty$. Formalizing this to series is not too hard: define $$a_i=2i-1\qquad\text{and}\qquad b_i=2i.$$ Then we can show $A_n=n^2$ while $B_n=n^2+n$ and the problem is essentially the same as how we treated $f(x)$ and $g(x)$ above.

This is not true. Suppose that $$A_n=1+n(n+1)$$ $$B_n=n(n+1)$$ Then $$\lim_{n\to\infty} \frac{1+n(n+1)}{n(n+1)}=1$$ But $$\lim_{n\to\infty} (n(n+1)-(1+n(n+1)))=-1\ne0$$ A valid pair of sequences that give these $A_n,B_n$ is $$a_i=\begin{cases}3&i=1\\2i&\text{otherwise}\end{cases}$$ $$b_i=2i$$