Proving that $\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}\right)$ has a limit $$x_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}$$
How can we prove that the sequence $(x_n)$ has a limit? I have to use the fact that an increasing sequence has a limit iff it is bounded from above. No more "advanced" tools can be used. It's obvious that this sequence is increasing, but I am having trouble finding a bound.

Alternately you can use 
$$x_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!} \leq x_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{(n-1)\cdot n}$$
and
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{(n-1)\cdot n}$$ is telescopic, since
$$\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1} \,.$$
