Does a function which expressed as a taylor series differentiable and/or continuous in the range of convergence Suppose that f(x) is infinitely differentiable function at a neighborhood of $0$, and that the radius of convergence of its taylor series around $0$, $\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n$, is 8.
Does it mean that $f(x)$ is continuous and/or differentiable at $(-8,8)?$ and why?
 A: No, it does not. For instance, I can take any power series $\sum_{k=0}^\infty a_k x^k$ with radius of convergence $8$, and then define
$$\begin{align*}f:&\mathbb R\to\mathbb R\\
&x\mapsto\begin{cases}\sum_{k=0}^\infty a_k x^k&x\in(-1,1)\\0&\textrm{otherwise.}\end{cases}\end{align*}$$
The Taylor expansion of this function around $0$ is just the given power series, but it only agrees with the power series within the interval $(-1,1)$, even though the power series has larger radius of convergence. But if $f$ actually agrees with its Taylor series on $(-8,8)$, in other words, it is analytic, then yes, it will be differentiable (even infinitely often) on the entire interval. But analyticity is a very strong condition, so you can't always assume it.
A: Relation between given $f$ function and its Taylor series can be tricky. It's famous example
$$f(x)=\begin{cases}
e^{-\frac{1}{x^2}}, & x \ne 0 \\
0, & x=0
\end{cases}
$$
which is infinitely differentiable with $f^{(n)}(0)=0, \forall n \in \mathbb{N}$. The Taylor series $\sum_{i=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n = 0$ converges on all $\mathbb{R}$, i.e. its radius of convergence is $R=\infty$ but coincides with function only at origin. Now we can take some function $g$ wich equal $f$ only at origin's some neighbourhood, but can be any type outside, for example not continuous.
So it it useful to have necessary and sufficient condition for given $\boldsymbol{f}$ function to be representable by its Taylor series on convergence interval $(-R,R)$, where $R$ is radius of convergence. One is following:
Taylor remainder in Maclaurin form $R_{n+1}=\left( \frac{x-a}{x-\xi} \right)^p\frac{(x-\xi)^{n+1}}{n!p}f^{(n+1)}(\xi)$ on given interval tends to $0$, where $p>0$, $\xi$ between $x$ and $a$.
