$\sum\limits_{n=0}^\infty\left(\dfrac{1}{n!}\right)$ converges

$\sum\limits_{n=0}^\infty\left(\dfrac{1}{n!}\right)$ converges since:

(1) $\sum\limits_{n=0}^\infty\left(\dfrac{1}{n!}\right) = \dfrac{1}{0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \dots + \dfrac{1}{n!} = $

$$= 1 + 1 + \frac{1}{1\times{2}} + \frac{1}{1\times{2}\times{3}} + \dots + \frac{1}{1\times{2}\times{3}\times\dots{n}} <$$

$$< 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{n-1}}$$    

(2)  Let $u = 1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dots + \dfrac{1}{2^{n-1}}$ 

(3)  $u - \dfrac{1}{2}u = \dfrac{1}{2}u = 1 - \dfrac{1}{2^n}$ so that $u = 2 - \dfrac{1}{2^{n-1}} < 2$

(4)  So it follows that $\sum\limits_{n=0}^\infty\left(\dfrac{1}{n!}\right) < 1+2=3$