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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 :-)

Fasal123
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  • Yes same, or rather in a symmetric form, a harmony sum $\frac{1}{2\cdot 1499} \left( \frac{1}{1499-r}+\frac{1}{1499+r}\right)$, $r$ from $-499$ to $499$ – King Tut Apr 01 '18 at 11:30
  • I note that wolfram alpha doesn't find an elegant resolution of the sum. – lulu Apr 01 '18 at 11:31
  • It was asked in my test, so I guess it must be correct. Plus, I expected the method to be kind of simple, just that I was missing it. – Fasal123 Apr 01 '18 at 11:33
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    Have you tried this form $$ \begin{array}{rcl} \frac{1}{1000\cdot 1998}&=&\frac{1}{998}(\frac{1}{1000}-\frac{1}{1998}) \ \frac{1}{1001\cdot 1997}&=&\frac{1}{996}(\frac{1}{1001}-\frac{1}{1997}) \ &\vdots& \end{array} $$ – Amin235 Apr 01 '18 at 11:37
  • @Amin235 I think I tried it, but it wasn't useful – Fasal123 Apr 01 '18 at 11:48
  • @DonAntonio No they indeed are true – Rohan Shinde Apr 01 '18 at 11:56
  • @Manthanein My bad: they are true...:). Deleting nonsensical comment. – DonAntonio Apr 01 '18 at 12:00
  • @DonAntonio $$\frac {1}{998}\left(\frac {1}{1000}-\frac {1}{1998}\right) =\frac {1}{998}\cdot \frac {\color{red}{998}}{1998\cdot 1000}$$ – Rohan Shinde Apr 01 '18 at 12:02
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    This question is missing context or other details: Please improve the question by providing additional context, which ideally includes the specific source of the problem, its wider motivation, your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Carl Mummert Apr 01 '18 at 14:53
  • Please see "How to ask a good question" at https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question for further advice. – Carl Mummert Apr 01 '18 at 14:53

1 Answers1

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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 $B\approx 4.62\cdot 10^{-4}$

Jim Haddocc
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