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?
 A: 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.
A: $$\frac{999}{1999}+\frac{(999)(998)}{(1999)(1998)}+....\frac{999!}{(1999)(1998)...(1001)}$$
$$\sum_{k=0}^{998}\frac{\frac{999!}{k!}}{\frac{1999!}{(1000+k)!}}$$
$$\frac{(999!)}{(1001)(1002)...(1999)}\sum_{k=0}^{998}\frac{(1000+k)!}{(k!)(1000!)}$$
Note that,
$$\binom{x}{y}=\frac{x!}{(y!)(x-y)!}$$
$$\frac{999!}{(1001)(1002)...(1999)}\sum_{k=0}^{998}\binom{1000+k}{1000}$$
By using Christmas stick identity (Also known as hockey stick identity) we get it's equal to,
$$\frac{(999!)(1999!)}{(1001)(1002)...(1999)(1001!)(998!)}$$
Which after simplification leads to
$$\frac{999}{1001}$$
