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 $n\to \infty$.
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$\begingroup$ It's a Taylor polynomial of $e^x$, so you can use the integral form of the remainder (this way you just get the answers below) $\endgroup$– Pietro MajerApr 16, 2016 at 15:37
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2 Answers
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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.
answered Apr 16, 2016 at 8:11
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According to Wolfram alpha it is related to the incomplete gamma function $\Gamma(a,x)$ and equals $$\frac{e^x\Gamma(n+1,x)}{n!}$$
