$\lim\limits_{n \to \infty}\left(1 + \dfrac{1}{n}\right)^n$ converges

$\lim\limits_{n \to \infty}\left(1 + \dfrac{1}{n}\right)^n$ converges since:

(1)  From the binomial theorem:  

$$\left(1+\frac{1}{n}\right)^n = \sum\limits_{k=0}^{n}{n \choose k}\left(\frac{1}{n}\right)^{k} = $$ $$=1 + \frac{n!}{(n-1)!}\left(\frac{1}{n}\right) + \frac{n!}{(n-2)!2!}\left(\frac{1}{n}\right)^2 + \frac{n!}{(n-3)!3!}\left(\frac{1}{n}\right)^3 + \dots + \frac{n!}{n!}\left(\frac{1}{n}\right)^n$$

$$=1 + 1 + \frac{n-1}{2!}\left(\frac{1}{n}\right) +  \frac{(n-1)(n-2)}{3!}\left(\frac{1}{n}\right)^2 + \dots + \frac{(n-1)(n-2)\dots(2)}{n!}\left(\frac{1}{n}\right)^{n-1}$$

$$=1 + 1 + \frac{1}{2!}\left(1 - \frac{1}{n}\right) + \frac{1}{3!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) + \dots +$$ $$+ \frac{1}{n!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\dots\left(1 - \frac{n-1}{n}\right)  \le \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{n!}$$

(2) From this result, we know that $\lim\limits_{n \to \infty}\sum\limits_{i=0}^n\left(\dfrac{1}{i!}\right)$ converges.