Simple Taylor's Expansion

According to Taylor's expansion of $e^x$ , $\displaystyle\sum _{ n=1 }^{ \infty }{ \frac { x^n }{ n! } } = e^x$ . Therefore for $x=1$ , we have: $\displaystyle\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n! } } = e$ => $1+ \displaystyle\sum _{ n=0 }^{ \infty }{ \frac { 1}{ n! } } = e$ => $\displaystyle\sum _{ n=0 }^{ \infty }{ \frac {1}{ n! } } = e-1= 1.718:)$

$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$

$x=1 \implies e=1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots$

$\implies e-1 =1+\frac{1}{2!}+\frac{1}{3!}+\cdots=\sum_{n=1}^{\infty}\frac{1}{n!}$

Therefore, $1000\sum_{n=1}^{\infty}\frac{1}{n!}=1000(e-1)=1000(1.718)=1718.$