= Solution
For <uniform convergence> to $k$ on $A$, for every $\eta>0$ there is $N$ such that $|k_n(x)-k(x)|<\eta$ for all $n\ge N$ and all $x\in A$. The bound $|\sin(nx)/\sqrt n|\le1/\sqrt n$ proves <uniform convergence> to zero, whereas its <derivative> is $\sqrt n\cos(nx)$. On an interval containing zero these <derivatives> diverge at zero, so they cannot converge uniformly. This supplies a counterexample to differentiating a uniformly convergent sequence without additional hypotheses.
For <uniform derivative convergence with an anchored value>, put $g=\lim k_n'$ and $a=\lim k_n(x_0)$. A uniform limit of continuous functions is continuous. The <fundamental theorem of calculus> gives
$$
k_n(x)=k_n(x_0)+\int_{x_0}^x k_n'(t)\,dt.
$$
Thus $k_n(x)\to k(x)=a+\int_{x_0}^x g(t)\,dt$, with error bounded by
$$
|k_n(x_0)-a|+|x-x_0|\sup_A|k_n'-g|.
$$
This is uniform on bounded subintervals, and on all of $A$ if it is bounded. The <fundamental theorem of calculus> now gives $k'=g$, so $k$ is continuously differentiable. For an unbounded $A$, no global <uniform convergence> of $k_n$ follows merely from the displayed hypotheses, but local <uniform convergence> and the claimed differentiability do.
For the series, use partial sums and differentiate their individual terms:
$$
\frac{d}{dx}\frac{x^n\sin(nx)}{n^3+1}
=\frac{n x^{n-1}\sin(nx)+n x^n\cos(nx)}{n^3+1}.
$$
For $|x|\le1$, their absolute values are at most $2n/(n^3+1)$, a summable bound. The <Weierstrass M-test> makes the <derivative> series uniformly convergent, while every partial sum is zero at $x=0$. The anchored-value result therefore proves \b[continuous differentiability on $(-1,1)$], with <derivative> equal to the displayed termwise <derivative> series.
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