Find $\sum_{n=1}^{\infty}\frac{1}{n!}$ 
Find $$\sum_{n=1}^{\infty}\frac{1}{n!}$$

All of the advice I've seen to compute this sum says to use the ratio test, but this is in a chapter BEFORE the ratio test, so the book wants me to solve this without the ratio test. My tools are the comparison test, the limit comparison test, and basic knowledge about geometric series and p-series.
I can't find anything to compare this function to. I wish I could type some more work, but I don't even know how to get started.
EDIT: Yes, I had a typo. It is an infinite sum.
 A: I will assume that you are taking about the infinite sum
Use induction to show that $1/n!<2^{-n}$ for all n>2
Then use the comparison test
A: $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$$
Put $x=1$. Therefore, the answer is $e-1$.
A: You say you want to find $\sum\limits_{n=0}^\infty\dfrac{1}{n!}$.  But then in comments you say you want to find out whether it converges or not.  So you're not expressing yourself clearly.
If you just want to find out whether it converges or not, you can use this comparison test:
$$
\frac{1}{n!} \le 2\cdot\frac{1}{2^n}.
$$
A: Since you mention $p$-series and comparison test:
First: $$\sum_{n=1}^{\infty}\frac{1}{n^2} = \sum_{n=2}^{\infty} \frac{1}{(n-1)^2}$$ is convergent because it is a $p$ series with $p=2$.
Second
$$
0\leq \frac{1}{n!} = \frac{1}{1}\frac{1}{2}\dots \frac{1}{n-1}\frac{1}{n} < \frac{1}{n(n-1)} < \frac{1}{(n-1)^2}
$$
for all $n\geq 2$. Hence by the comparison test $\sum_{n=2}^{\infty}\frac{1}{n!}$ is convergent, and so then is $\sum_{n=1}^{\infty}\frac{1}{n!}$.
