Question-163430

Question Number 163430 by amin96 last updated on 06/Jan/22

Answered by qaz last updated on 07/Jan/22

\frac{A}{B} = \frac{\underset{n = 1}{\sum^{999}} \frac{1}{(2 n - 1) (2 n)}}{\underset{n = 1}{\sum^{999}} \frac{1}{(999 + n) (1999 - n)}}

= \frac{\underset{n = 1}{\sum^{999}} (\frac{1}{2 n - 1} - \frac{1}{2 n}) = H_{1998} - \frac{1}{2} H_{999} - \frac{1}{2} H_{999} = H_{1998} - H_{999}}{\frac{1}{2998} \underset{n = 1}{\sum^{999}} (\frac{1}{999 + n} + \frac{1}{1999 - n}) = \frac{1}{1499} \underset{n = 1000}{\sum^{1998}} \frac{1}{n} = \frac{1}{1499} (H_{1998} - H_{999})}

= 1499

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