Taylor expansion of $\exp(-x)$: proof tail is positive

Question

The Taylor expansion of $\exp(-x)$ is $$e^{−} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} ....$$ What I want to prove is that $$ \frac{x^2}{2!} - \frac{x^3}{3!} .... $$ is greater than or equal to $0$.

I would really appreciate some help.

Answer 1

By Lagrange remainder, we have:

$$e^{-x}-1+x=\frac{e^{-\xi}}2x^2=\sum_{n=2}^\infty\frac{(-1)^n}{n!}x^n$$

for some $\xi$ between $0$ and $x$. However, $e^{-\xi}\ge0$ and $x^2\ge0$, and thus, we have

$$\sum_{n=2}^\infty\frac{(-1)^n}{n!}x^n\ge0$$

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This specific example can be seen as integrating $e^{-x}$ twice:

$$e^{-x}-1-x=\int_0^x\int_0^ye^{-t}~\mathrm dt~\mathrm dy\ge0$$

One can bound the integrand and show that if $e^{-t}\le M$ then we have

$$\int_0^x\int_0^ye^{-t}~\mathrm dt~\mathrm dy\le\int_0^x\int_0^yM~\mathrm dt~\mathrm dy=\frac M2x^2$$

and likewise for a lower bound of $\frac N2x^2$ where $e^{-t}\ge N$. By the intermediate value theorem we must have

$$\int_0^x\int_0^ye^{-t}~\mathrm dt~\mathrm dy=\frac{e^{-\xi}}2x^2$$

for some $\xi$ between $0$ and $x$, which are forced by the bounds on $t$.

Answer 2

Consider $g(x)=e^{-x}-(1-x)$.

Note that it is $0$ when $x=0$, and its second derivative is $e^{-x},$ which is always positive.

That is, $g(x)$ is convex (concave upward) and therefore non-negative; QED.