uniformly convergent sequence of differentiable functions, series of derivative of terms not convergent I am attempting to come up with a uniformly convergent sequence of differentiable functions $g_{n}:(0,1)\to \mathbb R$ such that the sequence $\{g_{n}'\}$ does not converge. 
I was thinking that $g_{n}(x)= \frac{\sin(nx + 3)} { \sqrt(n+1)}$ would work, however my friend does not think so. Why wouldn't this work? I am having trouble thinking of an example where this would be true.
 A: I think your example works. I'd like to simplify it though. Let's let $g_n(x) = \sin (nx)/\sqrt n, n = 1,2,\dots $ Clearly $|g_n(x)|\le 1/\sqrt n$ for all $x \in \mathbb {R},$ so $g_n \to 0$ uniformly on $\mathbb {R}.$
On the other hand, $g_n'(x) = \sqrt n \cos (nx).$ Claim: For every $x\in (0,1),$ $g'_n(x)\ge \sqrt {n/2}$ for infinitely many $n,$  and $g'_n(x)\le -\sqrt {n/2}$ for infinitely many $n.$ Thus the sequence $g'_n(x)$ oscillates unboundedly as $n\to \infty,$ hence certainly doesn't converge.
The claim follows from this lemma: If $A$ is an arc on the unit circle of arc length greater than $1,$ and $x\in (0,1),$ then $ e^{inx} \in A$ for infinitely many $n.$ Proof: The sequence $e^{i nx}$ marches around the circle infinitely many times in steps of fixed arc-length $x<1.$ Now remember what your parents taught you: You can't step over a puddle that's larger than your stride. Apply this to our sequence: Because the steps are smaller than the length of $A,$ we have to land in $A$ at least once every orbit. That's the idea, and you can make it perfectly rigorous.
To get the claim from the lemma, consider the arcs $A= \{e^{it}: -\pi/4 < t < \pi/4\}, B = \{e^{it}: 3\pi/4 < t < 5\pi/4\}.$ The lengths of these arcs each equal $\pi/2>1.$ Apply the lemma and the claim follows.
