= Solution
For $f$ sufficiently differentiable on the interval between $a$ and $x$, <Taylor's theorem> with its first $n$ terms reads
$$
\boxed{f(x)=\sum_{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a)^k+R_n(x),\qquad R_n(x)=\frac{f^{(n)}(\xi)}{n!}(x-a)^n,}
$$
where $\xi$ lies between $a$ and $x$. This is the <Taylor theorem with Lagrange remainder>. For example, $|f^{(n)}|\le M$ on that interval gives $|R_n(x)|\le M|x-a|^n/n!$. The finite formula does not assume that an infinite <Taylor series> converges to the function; that further conclusion requires the remainders to tend to zero.
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