prove $| \exp(x) - 1 - x - \frac{x^2}{2!} - \frac{x^3}{3!}| < \frac{e}{24}$

Question

prove $\displaystyle\left\lvert \exp(x) - 1 - x - \frac{x^2}{2!} - \frac{x^3}{3!}\right\rvert| < \frac{e}{24}$ $\forall x \in [-1,1]$

my attempt,

we defined $\exp(x) = \displaystyle \sum_{n=0}^{\infty} \dfrac{x^n}{n!}$ using this:

$| \exp(x) - 1 - x - \frac{x^2}{2!} - \frac{x^3}{3!}| = \left | \displaystyle \sum_{n=4}^\infty \dfrac{x^n}{n!} \right | \leq \displaystyle\sum_{n=4}^\infty \left |\dfrac{x^n}{n!} \right | $ since $|x| \leq 1 $ and $e = 2.718...$ $\displaystyle\sum_{n=4}^\infty \left |\dfrac{x^n}{n!} \right | \leq \displaystyle\sum_{n=4}^\infty \left |\dfrac{e^n}{n!} \right | < e^4/24 < e/24$

Is this a valid proof? The reason I ask is because we have just started Taylor Series in my analysis class and I don't see how I'm exactly using this at all in the proof

edit using maclaurin:

$\exp(x) = 1 + x + x^2/2! + x^3/3! + e^\theta x^4/4!$ for some $\theta \in (0,1)$, since $|x| \leq 1$

$$\dfrac{e^\theta}{4!}|x^4| \leq e^\theta/24 < e/24$$

is this correct?

Answer 1

Using Maclaurin formula: there's $\theta\in(0,1)$ such that $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{e^{\theta x}}{4!}x^4$$ and since $x\in[-1,1]$ then we have $$e^{\theta x}x^4\le e^{|\theta x|}\le e^1=e$$ and we deduce the desired result.