What is the closed form of
 $$\sum\_{k=0}^n \frac{x^k}{k!}$$
 as a function of $x$ and $n$?


Knowing that it converges to $e^x$ when $n\to \infty$.



Is there a closed form for $\sum_{k=0}^n \frac{x^k}{k!}$?
Denoting this by $f\_n(x)$, we get $f\_n'-f\_n=-x^n/n!$, solving this differential equation we have $(f\_n(x)e^{-x})'=-x^n e^{-x}$, thus taking into account initial condition $f\_n(0)=0$ we get integral representation $$f\_n(x)=e^x-\frac1{n!}e^x\int\_0^x t^n e^{-t}dt.
$$
This may be further rewritten as $$f\_n(x)=\frac1{n!}e^x\int\_x^\infty t^n e^{-t}dt,$$
this is what is called incomplete Gamma function, see joro's answer.


