How to find $F_{x}(x)$ for a probability distribution? In a problem I am trying to solve the probability is given by

$p(x)=\frac{5^{x}e^{-5}}{x!}$, for $x=0,1,2,3,...$

The question asks to find $\mu_{x}(t)$, $E(x)$, $var(x)$ and $F_{x}(x)$.

I have already found the moment generating function $\mu_{x}(t)$, the mean $E(x)$ and the variance $var(x)$. However I am not sure what $F_{x}(x)$ means. I guess that $F$ is the cumulative distribution function so $F_{x}(x)$ probably means $P(x\leq x)$. But $p(x)$ is a discrete distripution so am I supposed to evaluate $\sum_{x=0}^{x}p(x)$. But how to evaluate this sum since the upper limit is not known?
Stating that: $$F_X(x)=\sum_{n=0}^{\lfloor x\rfloor} p(n)=e^{-5}\sum_{n=0}^{\lfloor x\rfloor}\frac{5^n}{n!}$$
is all you can do here. There is no closed form for the RHS.
Note that
$$p(x)=\frac{5^{x}e^{-5}}{x!} \sim \text{Pois}(5)(x)$$
so the cumulative distribution is
$$\begin{align}
F_x(x) &amp;= \sum_{i=0}^{x}p(i)\\
&amp;=\sum_{i=0}^{x}\text{Pois}(\gamma)(i)\\
&amp;=\sum_{i=0}^{x} \frac{\gamma^{i}e^{-\gamma}}{i!}\\
&amp;=e^{-\gamma} \sum_{i=0}^{x} \frac{\gamma^{i}}{i!}\\
&amp;\le e^{-\gamma} \frac{(e\gamma)^x}{x^x} \quad \text{if } x \le \gamma
\end{align}$$