A smooth function that is not its Taylor series

1. By induction: $f=P_0\left(\frac1x\right)e^{-1/x^2}$ with $P_0=1$. If $f^{(n)}=P_n(1/x)e^{-1/x^2}$, then
$$f^{(n+1)}=\left(-\frac1{x^2}P_n'\!\left(\frac1x\right)+\frac2{x^3}P_n\!\left(\frac1x\right)\right)e^{-1/x^2}.$$
Writing $u=\frac1x$ this is $\bigl(-u^2P_n'(u)+2u^3P_n(u)\bigr)e^{-u^2}$, so $P_{n+1}(u)=2u^3P_n(u)-u^2P_n'(u)$ is a polynomial.

2. For every polynomial $P$, $P\!\left(\frac1x\right)e^{-1/x^2}\to0$ as $x\to0$ (put $u=\frac1x\to\pm\infty$). Suppose $f^{(n)}(0)=0$. Then
$$\frac{f^{(n)}(x)-f^{(n)}(0)}x=\frac1xP_n\!\left(\frac1x\right)e^{-1/x^2}\to0,$$
so $f^{(n+1)}(0)=0$. Induction on $n$.

3. The Taylor series of $f$ at $0$ is identically $0$, while $f(x)>0$ for every $x\ne0$. So $f$ is not equal to its Taylor series on any interval around $0$.

Remark. Smooth ($C^\infty$) does not imply analytic. This function is the standard building block for bump functions and partitions of unity in analysis.