Taylor series not converging, other example than $\exp(-1/x^2)$? The usual example for non-converging Taylor series is $g(x) = \exp(-1/x^2) \; \forall x \neq 0, g(0) = 0$: the Taylor series around $x=0$ is zero, but $g$ isn't zero for any $x \neq 0$.
What's not so nice about this example are the derivatives: 
\begin{align*}g'(x) &= \frac{2}{x^3}\exp(-1/x^2), \\ g''(x) &= \left(-\frac{6}{x^4}+\frac{4}{x^6}\right)\exp(-1/x^2), \\ \ldots 
\end{align*}
So obviously we have to calculate the derivative of $g$ in $x = 0$ by using the definition of the derivative. It is not possible to get it by "applying the rules" (e.g. $2\exp(-1/x^2)/x^3$ isn't defined for $x=0$).
The question is: Can there be a function $f\colon \mathbb{R}\rightarrow \mathbb{R}$ which has a Taylor series with radius of convergence $0$, but whose derivatives in the Taylor series can be calculated easily by just using the "usual" rules (derivatives of polynomials, product rule, chain rule, derivatives of $\exp, \sin, \ldots$)?
 A: Here is an example: 
$$f(x)=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-t^2-x^2t^4}dt,\qquad x\in\mathbb{R}.$$
The function $f(x)$ as well as all of its derivatives are well-defined at $x=0$. Moreover, one can compute them explicitly:
$$f^{(2k)}(0)=(-1)^k\frac{(4k)!}{4^{2k}k!},\qquad f^{(2k+1)}(0)=0.$$ 
However the radius of convergence of the Taylor series is $0$.
A: You are asking for $f$ to be not analytic, but such that its derivatives can be found using derivative rules (product, quotient, chain) starting from elementary functions. There are no such examples. 
The derivative rules apply only when the "building blocks" of the function are differentiable at the given point (and in case of the quotient, the denominator is not zero). The building blocks we start with (exponential, trigonometric, etc) are such that they are analytic everywhere where they are differentiable. And analyticity is also preserved by product, quotient and composition (for quotient: assuming the denominator is not zero). 
Summary: if the derivative rules apply, the analyticity is not lost. 
