Evaluation of $\sum^{\infty}_{n=0}\frac{999!\cdot n!}{(1001+n)!}$

Question

Evaluation of $\displaystyle \sum^{\infty}_{n=0}\frac{999!\cdot n!}{(1001+n)!}$

What I try :

$\displaystyle 1000\times 1001\sum^{\infty}_{n=0}\frac{1001! \cdot n!}{(1001+n)!}$

$\displaystyle =1001\cdot 10^3\sum^\infty_{n=0}\frac{1}{\binom{1001+n}{n}}$

$\displaystyle =1001\cdot 10^3\sum^{\infty}_{n=0}\int^1_0 x^{1001-n}\cdot (1-x)^ndx$

(Above using $\displaystyle \int^1_0 x^m\cdot (1-x)^ndx=\frac{1}{m+n+1}\cdot \frac{1}{\binom{m+n}{n}}$

How can I find value of above sum, Please have a look on this , Thanks

Answer 1

$$\sum^{\infty}_{n=0}\frac{999!\cdot n!}{(1001+n)!}=\frac{999!}{1001!}+\sum^{\infty}_{n=1}\frac{999!\cdot n!}{(1001+n)!} $$ We have : $$\frac{999!\cdot n!}{(1001+n)!}=\frac{999!n!}{1×2×3×...×n×(n+1)×...(n+1001)}=\frac{999!}{(n+1)...(1001+n)}$$

therfore: $$\sum^{\infty}_{n=0}\frac{999!\cdot n!}{(1001+n)!}=\frac{999!}{1001!}+\sum^{\infty}_{n=1}\frac{999!}{(n+1)...(1001+n)} $$

we have : $$\sum^{\infty}_{n=1}\frac{1}{(n+1)...(1001+n)}=\frac{1}{1000}\sum^{\infty}_{n=1}\left({\frac{1}{(n+1)...(1000+n)}-\frac{1}{(n+2)...(1001+n)}}\right)=\frac{1}{1000×1001!}$$

therfore: $$\sum^{\infty}_{n=0}\frac{999!\cdot n!}{(1001+n)!}=\frac{999!}{1001!}+\frac{999!}{1000×1001!}=\frac{1}{10^6}$$