The given sequence is: $ x_n = \frac{1}{1!} + \frac{1}{2!} + \dots + \frac{1}{n!} $
I want to first try showing that it is a cauchy sequence.
The definitions I have for cauchy sequence are
1)$\forall \epsilon>0 ,\, \exists N \in\Bbb{N} \,$ $\ni $ $\forall n,m>N \ $ , it follows that $|x_m-x_n|<\epsilon$
2)$\forall \epsilon>0 ,\, \exists N \in\Bbb{N} \,$ $\ni $ $\forall n,>N \, p\in \Bbb{N} \ $ , it follows that $|x_m-x_{n+p}|<\epsilon$
Using Monotone Convergence Theorem
$ x_n = \frac{1}{1!} + \frac{1}{2!} + \dots + \frac{1}{n!} $
and
$ x_{n+1} = \frac{1}{1!} + \frac{1}{2!} + \dots + \frac{1}{n!}+ + \frac{1}{(n+1)!}$
It shows $x_n$ is increasing
Since $\frac{1}{n!}\leq\frac{1}{2^{n-1}}$ (it can prove by induction $\forall n \in N$)
$ x_n = \frac{1}{1!} + \frac{1}{2!} + \dots + \frac{1}{n!} \leq \frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\dots+\frac{1}{2^{n-1}}=2-\frac{1}{2^{n-1}}<2$, $n\in N$
It shows $x_n$ is bounded above
By the monotone convergence theorem, $x_n$ both increasing and bounded above then it is convergent.