Convergence of general Taylor series

Question

For any $a,h \in \mathbb{R}$, how can we see the series, $\sum_{k=0}^{\infty} \dfrac{f^{(k)}(a)}{k!} h^k$ converges to $f(a+h)$?

Answer 1

Suppose that $f$ equals it's taylor series. Then $$f(x) = \sum_{k=0}^{\infty} \frac{f^k(a) (x-a)^k}{k!} $$ and then performing the substitution $x \to x+h$ followed by $x \to a$ yields the formula you ask about.

As others have pointed out there are convergence issues, in general.

Answer 2

A standard counterexample is $f(x) = \begin{cases} e^{-{1 \over x^2}}, & x \neq 0 \\ 0, & x=0 \end{cases}$. The Taylor coefficients around $x=0$ are all zero, but the function is clearly not zero.

(Note that this has nothing to do with the radius of convergence of the terms ${ f^{(k)}(a) \over k!}$, as in this case, all of these terms are zero.)