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Find the value of the sum $\frac{1}{1000*1998}+\frac{1}{1001*1997}+\cdots+\frac{1}{1998*1000}$

I attempted expressing it as a telescoping sum, but I don't know how to. Also, I was curious, is there a closed formula for the general sum $\frac{1}{xy}+\frac{1}{(x+1)(y-1)}+\cdots+\frac{1}{yx}$, for positive integers $x<y$ and if so, how is it derived?

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  • $\begingroup$ You can find a pretty good approximate formula for it $\approx \frac{\log(2)}{1500}$ good to $0.03\%$ (using either a Riemann sum or the asymptotic formula for the $n$th harmonic number) $\endgroup$
    – Winther
    May 28, 2018 at 17:25
  • $\begingroup$ You can write $1000\times1998$ or $1000\cdot1998.$ Using an asterisk for that purpose is a workaround for occasions when one is limited to the characters on the keyboard. $\endgroup$
    – Michael Hardy
    May 28, 2018 at 18:11

1 Answer 1

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This is $$\sum_{n=1000}^{1998}\frac1{n(2998-n)} =\frac1{2998}\sum_{n=1000}^{1998}\left(\frac1n+\frac1{2998-n}\right) =\frac1{1499}\left(H_{1998}-H_{999}\right)$$ where $H_n$ is the $n$-th harmonic number.

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