How to simplify series in a fraction Given
$$\\ A =\frac{1}{1 \cdot 2} + \frac{1}{3 \cdot 4} + \frac{1}{5 \cdot 6} + \ ... \ + \frac{1}{1997 \cdot 1998} \\ B =\frac{1}{1000 \cdot 1998} + \frac{1}{1001 \cdot 1997} + ... + \frac{1}{1998 \cdot 1000}$$
Simplify $\frac{A}{B}$
First of all, I broke down each fraction in A term to partial fractions:
$\frac{1}{1 \cdot 2} + \frac{1}{3 \cdot 4} + \frac{1}{5 \cdot 6} + \ ... \ + \frac{1}{1997 \cdot 1998} \\ = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \ ... \ + \frac{1}{1997} - \frac{1}{1998}$
At this point, I noticed that the series of the partial fractions can be formed in Catalan Identity

Catalan Identity: $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ ... \ + \frac{1}{2n - 1} - \frac{1}{2n} = \frac{1}{n + 1} + \frac{1}{n + 2} + \ ... \ + \frac{1}{2n}$

Then I applied this identity to the series:
$A = \frac{1}{1000} + \frac{1}{1001} + \ ... \ + \frac{1}{1998}$
Subtitute it to A, I get:
$\frac{A}{B} = \frac{\frac{1}{1000} + \frac{1}{1001} + \ ... \ + \frac{1}{1998}}{\frac{1}{1000 \cdot 1998} + \frac{1}{1001 \cdot 1997} + ... + \frac{1}{1998 \cdot 1000}}$
That's how far I got. I'm stucked at this point. Can anybody show me how to finish this problem?

For any $a \neq 0$ and $b \neq 0$, you have
$$\frac{a + b}{ab} = \frac{1}{a} + \frac{1}{b} \tag{1}\label{eq1A}$$
Note the sum of each of the $2$ factors in the denominators of the summation in $B$ is the same, e.g., $2998 = 1000 + 1998 = 1001 + 1997 = \ldots$. Using this, along with \eqref{eq1A}, you have
$$\begin{equation}\begin{aligned}
2998\sum_{i=0}^{998}\frac{1}{(1000 + i)(1998 - i)} & = \sum_{i=0}^{998}\left(\frac{1}{1000 + i} + \frac{1}{1998 - i}\right) \\
& = \sum_{i=0}^{998}\left(\frac{1}{1000 + i}\right) + \sum_{i=0}^{998}\left(\frac{1}{1998 - i}\right) \\
& = 2\sum_{i=0}^{998}\left(\frac{1}{1000 + i}\right)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
The last line above comes from the second summation being of the same values as the first one, just in the opposite order. You now get
$$\sum_{i=0}^{998}\frac{1}{(1000 + i)(1998 - i)} = \frac{1}{1499}\sum_{i=0}^{998}\left(\frac{1}{1000 + i}\right) \tag{3}\label{eq3A}$$
Using your equation and \eqref{eq3A}, you thus have
$$\begin{equation}\begin{aligned}
\frac{A}{B} & = \frac{\frac{1}{1000} + \frac{1}{1001} + \ ... \ + \frac{1}{1998}}{\frac{1}{1000 \cdot 1998} + \frac{1}{1001 \cdot 1997} + ... + \frac{1}{1998 \cdot 1000}} \\
& = \frac{\sum_{i=0}^{998}\left(\frac{1}{1000 + i}\right)}{\frac{1}{1499}\sum_{i=0}^{998}\left(\frac{1}{1000 + i}\right)} \\
& = \frac{1}{\frac{1}{1499}} \\
& = 1499
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
