Define what it means for a sequence of functions $k_{n}: A \rightarrow \mathbb{R}, n=1,2, \ldots$, to converge uniformly on an interval $A \subset \mathbb{R}$.

By considering the functions $k_{n}(x)=\frac{\sin (n x)}{\sqrt{n}}$, or otherwise, show that uniform convergence of a sequence of differentiable functions does not imply uniform convergence of their derivatives.

Now suppose $k_{n}(x)$ is continuously differentiable on $A$ for each $n$, that $k_{n}\left(x_{0}\right)$ converges as $n \rightarrow \infty$ for some $x_{0} \in A$, and moreover that the derivatives $k_{n}^{\prime}(x)$ converge uniformly on $A$. Prove that $k_{n}(x)$ converges to a continuously differentiable function $k(x)$ on $A$, and that

$$k^{\prime}(x)=\lim _{n \rightarrow \infty} k_{n}^{\prime}(x)$$

Hence, or otherwise, prove that the function

$$\sum_{n=1}^{\infty} \frac{x^{n} \sin (n x)}{n^{3}+1}$$

is continuously differentiable on $(-1,1)$.