Show that $1 \le \sum_{n=1}^\infty\frac{1}{n!} \le 2$

Question

How can I prove that $\sum_{n=1}^\infty\frac{1}{n!}$ is limited this way $1 \le \sum_{n=1}^\infty\frac{1}{n!} \le 2$. I know that the lower bound is obvious because the sequence of the partial sums is monotone in a crescent way and the first term is $\frac{1}{1!}=1$, for the upper bound I think that maybe $\lim s_n = 2$ with $(s_n)$ being the sequence of partial sums of $\sum_{n=1}^\infty\frac{1}{n!}$, since this sequence is convergent by the Ratio Test, but how can I prove this?

Answer 1

Note that for $n\geq1$ we have that

$$2^{n-1}\leq n!$$

so

$$\frac{1}{2^{n-1}}\geq\frac{1}{n!}$$

and then

$$\sum_{n=0}^\infty\frac{1}{2^n}\geq\sum_{n=1}^\infty\frac{1}{n!}.$$