 Solution

Taylor's theorem gives a finite Taylor polynomial of degree $n$, consisting of the first $n+1$ terms, and a remainder:In the Taylor theorem with Lagrange remainder form, assuming the required derivatives exist on an interval containing $a$ and $x$,for some $\xi$ strictly between $a$ and $x$ when $x\ne a$; at $x=a$ the remainder is zero. In particular, if $|f^{(n+1)}|\le M$ on that segment, then $|R_{n+1}(x)|\le M|x-a{|}^{n+1}/(n+1)!$. This is a finite expansion; it does not assume equality with an infinite Taylor series.