How could I prove that the following is greater than or equal to $0$? $$\sum_{n=2}^\infty \frac{(-x)^n}{n!}$$
And is this series equal to:$$\frac{e^{-x}}{2}x^2$$ if yes, why?
Question related to: Taylor expansion of $\exp(-x)$: proof tail is positive
$$\sum_{n\geq 2}\frac{(-x)^n}{n!}=\sum_{n\geq 0}\frac{(-x)^n}{n!}-\sum_{n=0}^{1}\frac{(-x)^n}{n!} = e^{-x}-1+x $$ so your claim is equivalent to $e^{-x}\geq 1-x$ or $e^{z}\geq 1+z$, which holds by the convexity of $e^z$: the graph of $e^z$ lies above the graph of the tangent line at the origin.