How to solve an expression where series are in a fraction

Question

Given two equations: $$A = \frac{1}{1\times 2}+\frac{1}{3\times 4} +\frac{1}{5\times 6}+ ... + \frac{1}{1997\times 1998}$$ $$B = \frac{1}{1000 \times 1998}+\frac{1}{1001 \times 1997} + ... + \frac{1}{1998 \times 1000}$$

Find the value of $\frac{A}{B}$

. . .

I found out that $\frac{1}{1\times 2}+\frac{1}{3\times 4} +\frac{1}{5\times 6}+ ... + \frac{1}{1997\times 1998}$ is equal to $\frac{1}{1} - \frac{1}{2} +\frac{1}{3} -\frac{1}{4} + \frac{1}{5} - \frac{1}{6} + ... + \frac{1}{1997} - \frac{1}{1998}$

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and since $1 - \frac{1}{2} +\frac{1}{3} -\frac{1}{4} + \frac{1}{5} - \frac{1}{6} + ...+ \frac{1}{2n - 1} - \frac{1}{2n}$ equals $\frac{1}{n + 1} +\frac{1}{n + 2} + \frac{1}{n + 3} + ... + \frac{1}{2n}$

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Then $\frac{1}{1} - \frac{1}{2} +\frac{1}{3} -\frac{1}{4} + \frac{1}{5} - \frac{1}{6} + ... + \frac{1}{1997} - \frac{1}{1998}$ is equal to $\frac{1}{1000} +\frac{1}{1001} + \frac{1}{1002} + ... + \frac{1}{1998}$

. . .

But how do I find the value of $\frac{A}{B}$? Can anyone show me a hint?

Answer 1

HINT

We have that

$$A=\sum_{k=1}^{1997} \frac1{k(k+1)}, \quad B=\sum_{k=1000}^{1998} \frac1{k(2998-k)}$$

then use

$$\frac1{k(k+1)}= \frac1k -\frac1{k+1}$$

$$\frac1{k(2998-k)}= \frac1{2998k} +\frac1{2998(2998-k)}=\frac1{2998}\left( \frac1{k} +\frac1{(2998-k)}\right)$$