P-n-1-999-1-2n-1-2n-Q-n-1-999-1-999-n-1999-n-P-Q-

Question Number 66910 by naka3546 last updated on 20/Aug/19

P = \underset{n = 1}{\sum^{999}} \frac{1}{(2 n - 1) (2 n)}

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

\frac{P}{Q} = ?

Commented by naka3546 last updated on 20/Aug/19

I s i t c o r r e c t i f t h e a n s w e r 1499 ?