Ratio of series $A$ and $B$ is an integer.

Question

Let $$ A= {1\over 1\cdot 2}+\color{red}{1\over 3\cdot 4}+...+ {1\over 1997\cdot 1998}$$

and $$B= {1\over 1000\cdot 1998}+{1\over 1001\cdot 1997}+...+ {1\over 1997\cdot 1001}+{1\over 1998\cdot 1000}$$

Prove that $A\over B$ is an integer.

I could only find that $$ A= 1-{1\over 1998}$$ using standard trick ${1\over x(x+1)} = {1\over x}-{1\over x+1}$. But I could not find answer for the second one. It is supposed to be a task for 15 years old (Romanian) children!

Edit I write it down wrong, so the $A$ is not correctly calculated. And it is different $A$ as in suggested duplicate.

That this is supposed to be a task for 15 year old children shouldn't mean anything. It's all about context: where the problem is from, and which 15 year old children are expected to solve this problem, for instance. If it's part of a contest or exam meant to find basically the 10-20 best in a country, it's very much appropriate for the age. So, that makes me curious: where is this problem from? — Arthur, Sep 21 '18 at 10:22
$$\sum_{r=1000}^{1998}\dfrac1{r(2098-r)}=\dfrac1{2098}\sum_{r=1000}^{1998}\left(\dfrac1r+\dfrac1{2098-r}\right)=\dfrac2{2098}\sum_{r=1000}^{1998}\dfrac1r$$ — lab bhattacharjee, Sep 21 '18 at 10:25
https://math.meta.stackexchange.com/questions/16774/contest-problem-policy — Theo Bendit, Sep 21 '18 at 11:03
Possible duplicate of How to solve $\frac{1}{1000.1998}+\frac{1}{1001.1997}+\cdots+\frac{1}{1998.1000}$ — Jaap Scherphuis, Sep 21 '18 at 12:34

Oleg567 · Accepted Answer · 2018-09-22T17:49:48.997

When denote $$ H(n) = \sum_{k=1}^n \dfrac{1}{k}, $$ then $A$ could be expressed as $$ A = \left(\dfrac{1}{1}-\dfrac{1}{2}\right) + \left(\dfrac{1}{3}-\dfrac{1}{4}\right) + \ldots + \left(\dfrac{1}{1997}-\dfrac{1}{1998}\right) \\ = H(1998) - 2\left(\dfrac{1}{2}+\dfrac{1}{4}+\ldots+\dfrac{1}{1998}\right)\\ = H(1998) - H(999). $$

Based on lab bhattacharjee's comment, $B$ could be expressed as $$ B = \sum_{r=1000}^{1998}\dfrac1{r(2998-r)}\\=\dfrac1{2998}\sum_{r=1000}^{1998}\left(\dfrac1r+\dfrac1{2998-r}\right)\\=\dfrac2{2998}\sum_{r=1000}^{1998}\dfrac1r\\ = \dfrac{1}{1499} \Bigl(H(1998)-H(999)\Bigr). $$

Therefore $\dfrac{A}{B}=1499$.

Ratio of series $A$ and $B$ is an integer.

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