How to solve $\frac{1}{1000.1998}+\frac{1}{1001.1997}+\cdots+\frac{1}{1998.1000}$

The question is:

If $$A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\cdots+\frac{1}{1997.1998}$$ and $$B=\frac{1}{1000.1998}+\frac{1}{1001.1997}+\cdots+\frac{1}{1998.1000}$$ then what is the value of $\frac{A}{B}$?

I could figure out that $A=\frac{1997}{1998}$, but I have no idea how to proceed with $B$. Could somebody help?

Thanks for any help :-)

I could find a very precise approximate value for $B$. The sum can be written as $$B=\sum_{k=0}^{998}\frac{1}{(1000+k)(1998-k)}$$ $$=\frac{1}{2998}\sum_{k=0}^{998}\frac{1}{(1000+k)}+\frac{1}{(1998-k)}$$ $$=\frac{2}{2998}\sum_{k=0}^{998}\frac{1}{(1000+k)}$$ $$=\frac{2}{2998}(H_{1998}-H_{999})$$

Where $H_n=\sum_{i=1}^n \frac{1}{n}$. For large $n$, $H_n$ can be approximated as $H_n=\ln n$, so that in the last step we can approximately write

$$\frac{2}{2998}(H_{1998}-H_{999})\approx \frac{2}{2998}\ln \frac{1998}{999}$$ Maple tells us the error in this approximation is about $0.11\%$, which is quite small indeed. Thus $B\approx 4.62\cdot 10^{-4}$