Hagen answered a specific part of your question, but I see that your question is how to understand how to do this in general. So, that is why I am answering the question. Please let me know if there are parts that are not clear.
In general, the cumulative distribution function is defined by
$$F(x) = P(X \leq x).$$
You already have $p(x) = P(X = x)$. So, if you want to find $F(n) = P(X \leq n)$, you just add up all of the values of $p(x)$ for all $x \leq n$. That is
F(n) &= P(X \leq n) \\
&= P(X = 0) + P(X = 1) + P(X = 2) + \cdots + P(X = n) \\
&= p(0) + p(1) + p(2) + \cdots + p(n).
The above formula assumes that $n$ is an integer. But, if $n$ were not an integer, you would just add up all values of $p(x)$ for $x \leq n$, just as I said. For example,
$$F(3.14159) = p(0) + p(1) + p(2) + p(3).$$
Now, if you want to find a nice formula for $F(x)$ for this specific problem, see Hagen's answer. In general, nice formulas may or may not be possible, and the techniques for finding them would vary hugely from problem to problem.
For this specific problem, note, for any $x$ in $[0, 1)$, $F(x)$ will have the same value, $p(0)$. And, for any $x$ in $[1, 2)$, $F(x) = p(0) + p(1)$. And, for any $x$ in $[2, 3)$, $F(x) = p(0) + p(1) + p(2)$. And so on. So, $F(x)$, as a function, has an infinite number of jump discontinuities. The main thing here is, for any $x$ value, hopefully you now understand how to find $F(x)$.