Chapter 10: Infinite Sequences and Series - Section 10.10 - The Binomial Series and Applications of Taylor Series - Exercises 10.10 - Page 633: 41

$e=1+1+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...$

As we know that $e^x=1+x+\dfrac{x}{2!}-.....+\dfrac{x^{n}}{(n)!}$ Then, we have $e^{(1)}=1+1+\dfrac{1}{2!}+.....+\dfrac{1}{n!}$ Hence, $e=1+1+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...$