In one of the problems I am trying to solve, it basically narrowed down to finding the sum $$\sum^{n=c}_{n=0}\frac{x^n}{n!}$$ which is the partial sum of the Maclaurin series for $e^x$.
Wolfram | Alpha gave the output $$\frac{e^x\Gamma(c+1,x)}{\Gamma(c+1)}$$ But I am unsure how to derive this. All my usual summation evaluation tricks are not helpful and I am clueless.
Help is appreciated!
Finally got to doing this and got some help from aops xd
Anyhow derive the result by starting with the incomplete gamma function
$$e^x\Gamma(c+1, x) = \int_x^\infty \underbrace{t^c}_u \cdot \underbrace{e^{x-t}}_v \text{ d}t$$ Repeatedly integrating by parts and noting that $c$ is finite and an integer gives us \begin{align*} e^x\Gamma(c+1, x) &= x^c + c\cdot x^{c-1} + c(c-1)\cdot x^{c-2} + \cdots + c(c-1)\cdots 2\cdot x^1 + c! \cdot x^0 \\ &= c! \cdot \sum_{n=0}^c \frac{x^n}{n!} \\ \implies \frac{e^x\Gamma(c+1, x)}{c!} &= \sum_{n=0}^c \frac{x^n}{n!} \end{align*}
The series starting from $0$ spits out terms in reverse order compared to the integration by parts.
Using the gamma function gives the desired result $$\frac{e^x\Gamma(c+1, x)}{\Gamma(c+1)} = \sum_{n=0}^c \frac{x^n}{n!}$$
$$\begin{align*}\sum^{c}_{n=0}\frac{x^n}{n!}&=\sum^{\infty}_{n=0}\frac{x^n}{n!}-\sum^{\infty}_{n=c+1}\frac{x^n}{n!}\\ &=\sum^{\infty}_{n=0}\frac{x^n}{n!}-\sum^{\infty}_{n=0}\frac{x^{n+c+1}}{(n+c+1)!}\\ &=e^x-e^x\left(1-\frac{\Gamma(c+1,x)}{\Gamma(c+1)}\right) \\ &=e^x\frac{\Gamma(c+1,x)}{\Gamma(c+1)} \end{align*}$$ Here $\Gamma(a,x)$ is the Incomplete gamma function.
\Gammato produce $\Gamma$. $\endgroup$