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.
 A: 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
$$\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1} \,.$$
A: I know you may not be able to use this for the task at hand, but I couldn't resist pointing out is really good to know in the long run: 
The infinite sum given by $x_n$ defines (is one representation of) the mathematical constant e. That is,
$$ \lim_{n\to \infty} x_n\,=\,\lim_{n\to \infty} \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!} + ... \,=\,\sum_{n=0}^\infty \frac{1}{n!} \,= \,e.$$ 
So, indeed, as demonstrated by lab bhattacharjee, $$2.5 \;\lt\; \lim x_n = e \;\lt \;3$$
A: $x_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}<1+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n-1}}<1+\sum_{k=0}^\infty\frac{1}{2^k}$
A: $x_n>1+1+\frac12=2.5$
Nowas David Mitra suggested,
$$\frac1{3!}=\frac1{1\cdot2\cdot3}<\frac1{2\cdot2}=\frac 1{2^2},$$
$$\frac1{4!}=\frac1{1\cdot2\cdot3\cdot4}<\frac1{2\cdot2\cdot2}=\frac 1{2^3}$$ 
$$\frac1{5!}=\frac1{1\cdot2\cdot3\cdot4\cdot5}<\frac1{2\cdot2\cdot2\cdot2}=\frac 1{2^4}$$ 
and so on.
Hence $$x_n<1+1+\frac12+\frac1{2^2}+\frac1{2^3}+\cdots< 1+\sum_{0\le r<\infty}\frac1{2^r}=1+\frac1{1-\frac12}=3$$
