 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 $|\sin (nx)/\sqrt{n}|\le 1/\sqrt{n}$ 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 $g=\lim k_n^{\prime }$ and $a=\lim k_n(x_0)$. A uniform limit of continuous functions is continuous. The fundamental theorem of calculus givesThus $k_n(x)\to k(x)=a+{\int }_{x_0}^{x}g(t)dt$, with error bounded byThis is uniform on bounded subintervals, and on all of $A$ if it is bounded. The fundamental theorem of calculus now gives $k^{\prime }=g$, 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:For $|x|\le 1$, 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 continuous differentiability on $(-1,1)$, with derivative equal to the displayed termwise derivative series.