Taylor's Inequality. If $|f^{n + 1}(x)| \le M$ for $|x - a|\le d$, then the remainder $R_n(x)$ of the Taylor series satisfies the inequality$$|R_n(x)| \le {M\over{(n + 1)!}}|x - a|^{n + 1} \text{ for }|x - a| \le d.$$
I have a few questions surrounding Taylor's inequality.
The proof follows immediately from the proof of Taylor's theorem, which I am guessing is what you want explained. Recall that $f'$ is the best linear approximation for $f$ in the sense that $f(x) \approx f(c) + f'(c)(x - c)$ for $x \approx c$. Taylor's theorem is basically an extension of this idea. Suppose we want to approximate a function near $c$ by a degree $n$ polynomial $$p_n(x) = a_1 + a_1(x - c) + a_2(x - c)^2 + \cdots + a_n(x - c)^n.$$ How a function changes is determined by its derivatives. If $x$ is very close to $c$, then since the derivatives are continuous, $f^{(k)}(x)$ is very close to $f^{(k)}(c)$. So you can imagine that if we make $p^{(k)}_n(c) = f^{(k)}(c)$, then $p_n$ will be a pretty good approximation of $f$ near $c$. If you $k$ times differentiate $p_n$ and plug in $c$, you get $k!a_k$. So we want to choose $a_k = f^{(k)}(c)/k!$ (i.e. the Taylor coefficient).
As for why we should care, the usefulness comes from the fact that the theorem gives you a way to approximate functions with polynomials, which are about as well-behaved and easy to understand as you could hope for. IIRC, if you grab a calculator and type in something like $\sin(27)$, the calculator is actually using these Taylor polynomials to compute the rational approximation it gives you. Also, the Euler identity $e^{i\theta} = \cos\theta + i\sin\theta$ follows by substituting $i\theta$ into the Taylor series for $e^x$, $\cos x$, and $\sin x$.
