 Solution

The finite form of Taylor's theorem with the first $n$ terms iswhere $\xi$ lies strictly between $a$ and $x$ for $x\ne a$. A sufficient hypothesis is that $f$ has continuous derivatives through order $n-1$ on the interval joining the endpoints and an $n$th derivative in its interior. The displayed polynomial has degree at most $n-1$; its Taylor remainder uses the $n$th derivative, not the $(n+1)$st. At $x=a$ the remainder is zero. This is a finite Taylor polynomial identity and does not require an infinite Taylor series to converge.