$f^{(n)}(x)$ exist $\forall x \in I$, an interval in $\mathbb{R}$ but $f^{(n+1)}(x)$ does not exist for any $x \in I$

Question

$f'(x)$, $f''(x)$, $f^{(3)}(x)$, ..., $f^{(n)}(x)$ all exist $\forall x \in I$, an interval in $\mathbb{R}$ but $f^{(n+1)}(x)$ does not exist for any $x \in I$

Can such a function exist?

In words: Does there exist a function $f$ that is $n$-times differentiable on an interval $I$ but is not $(n+1)$-times differentiable anywhere on $I$ (i.e. $(n+1)$-differentiable nowhere on $I$)?

I think it can't but can't think of the reason why. I can construct a function with countably infinite points in I where the function is only differentiable n times, but, it seems impossible to have every point in the interval, I, have this property.

Answer 1

Yes. Take the Weierstrass function $f$. It is differentiable nowhere. If $F$ is a primitive of $f$ (which exists, since $f$ is continuous), then $F'$ exists, but $F''$ exists nowhere.

Answer 2

Take any continuous, nowhere differentiable function (e.g. Weierstrass function). Then integrate it $n$ times. First time you can do it, as the function is continuous. Later, you are integrating differentiable, so continuous, functions.