The sum of $\frac{999}{1999} + \frac{999}{1999} \frac{998}{1998} + ...+ \frac{999}{1999} \frac{998}{1998}...\frac{1}{1001}$? Ive tried to represent $\frac{999}{1999}$ as $1 - \frac{1000}{1999}$ and so on, but it didnt lead me anywhere. I also tried to represent it as $\frac{999!}{1999!} \cdot \frac{1998!}{998!} + \frac{999!}{1999!} \cdot \frac{1997!}{997!} + \dots + \frac{999!}{1999!} \cdot \frac{1002!}{2!} + \frac{999!}{1999!} \cdot \frac{1001!}{1!} + \frac{999!}{1999!} \cdot \frac{1000!}{0!}$. Basically, it means that all that's left to do is calculate what $ \frac{1998!}{998!}+\frac{1997!}{997!} + \dots + \frac{1000!}{1!}$ is, but i am stuck and don't have any idea how to do that. Is there any other way to do it faster?

It becomes easier if the terms are added from right to left.
The sum of the last two terms is
$$
\frac{999}{1999} \frac{998}{1998} \cdots \frac{3}{1003}\frac{2}{1002}\left(1 + \frac{1}{1001}\right) = \frac{999}{1999} \frac{998}{1998} \cdots \frac{3}{1003} \cdot \frac{2}{1001} \, .
$$
The sum of the last three terms is therefore
$$
\frac{999}{1999} \frac{998}{1998} \cdots \frac{4}{1004}\frac{3}{1003}\left( 1 + \frac{2}{1001}\right) =
\frac{999}{1999} \frac{998}{1998} \cdots \frac{4}{1004} \cdot \frac{3}{1001} \, .
$$
Continuing in this way one finally gets for the complete sum
$$
\frac{999}{1999} \left( 1 + \frac{998}{1001}\right) = \frac{999}{1001} \, .
$$

Another option is to write the sum as
$$
 \sum_{k=1}^{999} \frac{999! (1999-k)!}{(999-k)! 1999!}
= \frac{1}{\binom{1999}{1000}} \sum_{k=1}^{999} \binom{1999-k}{1000} = \frac{\binom{1999}{1001}}{\binom{1999}{1000}}
= \frac{999}{1001} \, ,
$$
using the the hockey-stick identity.
