# Calculating a constant given a probability density function

I have a probability distribution as follows, where X denotes the number of accidents $$P(X=i) = K \cdot \frac{2^i}{i!}, \quad \quad i=0,1,2,...$$ and I need to find out what the value of the positive constant K is. Usually to calculate a constant in the PDF, I integrate the function and equal it to $1$, so I can calculate the constant. But in this case, how can I integrate the function? I don't think it's possible since there's the $i!$, so maybe it has something to do with the $E(X)$ instead? So I guess it is something like $$E(X) = \sum_{i=0}^\infty i\cdot K \cdot \frac{2^i}{i!}$$ I'm not sure where to go on from here though. Also, the second part asks for the probability of having 3 or more accidents, which I believe means calculate $P(X \ge 3)$. How do I calculate this? I don't think it should have anything to do with gamma functions.$\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ I'd really appreciate your help. Thanks in advance!

All you need to do is make sure that $\sum_n P(X = n) = 1$, that is
$$1 = \sum_n k \frac{2^n}{n!} = k\sum_n \frac{2^n}{n!} = ke^2$$
so $k = e^{-2}$. In this last step I used
$$e^x = \sum_n \frac{x^n}{n!}$$