For $x>0$ find $\sum_{k=0}^\infty\frac{x^ke^{-x}}{k!}$

Question

I've been stuck on this question for some time now. I've tried using the partial sums of the series to find an equation $S_n$, which can, in turn, find the $n^{th}$ partial sum and then find the limit of $S_n$ as $k$ approaches infinity. However, I've been unable to find $S_n$. Is there perhaps a better way to solve the problem?

Answer 1

$$\sum_{k=0}^\infty\frac{x^ke^{-x}}{k!} = e^{-x}\sum_{k=0}^\infty\frac{x^k}{k!}$$ Note that the Taylor series of $e^x$ centered at $a$ is: $$e^a + \frac{e^a(x-a)}{1} + \frac{e^{2a}(x-a)^2}{2!} + \ ...$$ But, what is the Taylor expansion of $e^x$ centered at $a = 0$?

Answer 2

$$ \sum_{k=0}^\infty\frac{x^ke^{-x}}{k!}=e^{-x}\sum_{k=0}^\infty\frac{x^k}{k!}=1 $$