It is well-known that the function

$$f(x) = \begin{cases} e^{-1/x^2}, \mbox{if } x \ne 0 \\ 0, \mbox{if } x = 0\end{cases}$$

is smooth everywhere, yet not analytic at $x = 0$. In particular, its Taylor series exists there, but it equals $0 + 0x + 0x^2 + 0x^3 + ... = 0$, so while it has radius of convergence $\infty$, it is not equal to $f$ even in a tiny neighborhood of $0$.

There is also a function

$$f(x) = \sum_{n=0}^{\infty} e^{-\sqrt{2^n}} \cos(2^n x)$$

which is smooth everywhere (that is, $C^{\infty}$) yet analytic *nowhere*. In particular, the Taylor series at every point has radius of convergence $0$. In fact, "most" smooth functions are not analytic.

But this gets me wondering. Could there exist some function which is smooth everywhere, analytic nowhere, yet its Taylor series at any point has nonzero radius of convergence, and so converges to *something*, but that something is not the function, not even in a tiny neighborhood about the point of expansion? If yes, what is an example of such a function? If no, what is the proof that such a thing is impossible? And also, if no, what sort of restrictions exist on the convergence of the T.s.? At how many/what distribution of points can it converge to something which is not the function? I note that if we multiply together the two functions just given above, we have another smooth-everywhere, analytic-nowhere function, but this time at $0$ we have a convergent Taylor series (the same zero series as before -- just use the generalized Leibniz rule) which doesn't converge to the function in even a tiny neighborhood of $0$.

**EDIT** (Dec 31, 2013): With some Googling I came across a post to mathoverflow:

https://mathoverflow.net/a/81465

The Taylor series of the Fabius function at any dyadic rational actually has infinite radius of convergence (only finitely many terms are nonzero) but does not represent the function on any interval.

So it seems it is possible to have a function whose Taylor series converges to "the wrong thing" at a dense set of expansion points. But it still doesn't answer the question of whether that is possible for *all* expansion points on the *entire* real line.