Answer
$e=1+1+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...$
Work Step by Step
Since, $e^x=1+x+\dfrac{x}{2!}-.....+\dfrac{x^{n}}{(n)!}$
Now, $e^1=1+1+\dfrac{1}{2!}+.....+\dfrac{1}{(n)!}$
Thus,
$e=1+1+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...$
