$\def\R{\mathbf{R}}$ Let $f:(-1,1)\to\R$ be infinitely differentiable. Then it has a taylor series expansion $T(x)=\sum_{n=0}^{\infty}{a_nx^n}$ where $a_n=f^{(n)}(0)/n!$. Let's say we're given that $T$ converges pointwise to $f$ on the interval $[0,1)$. What can we then say about its converge elsewhere?
We can deduce that $T$ converges to something on $(-1,1)$ absolutely, and uniformly on any compact subset of $(-1,1)$. But the lingering question is, must $T$ necessarily converge to $f$ on all of $(-1,1)$? In other words, is $f$ real analytic?
Not necessarily. Take the function$$\begin{array}{rccc}f\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}e^{-1/x^2}&\text{ if }x<0\\0&\text{ otherwise.}\end{cases}\end{array}$$Then the Taylor series of $f$ at $0$ is the null series, which converges to $f(x)$ for every $x\in[0,\infty)$, but which converges to $0\ne f(x)$ for every $x<0$.