 Solution

Taylor's theorem expands about $a$ in powers of $h=x-a$. With the usual differentiability assumptions, its finite form with Lagrange remainder iswhere $\xi$ lies between $a$ and $x$. If the remainder tends to zero as $n\to \infty$, this yields the Taylor series $f(x)={\sum }_{k\ge 0}{f}^{(k)}(a)(x-a{)}^k/k!$. Smoothness alone does not guarantee that infinite-series equality.