Show that $\sum_{n=1}^{\infty}\frac{sin(x^{n})}{n!}$ converges uniformly for $x \in \mathbb R$ to a $C^{1}$ function $f:\mathbb R \rightarrow \mathbb R$, compute an expression for the derivative.
My attempt: For uniform convergence, It is clear that $|\sum_{n=1}^{\infty}\frac{sin(x^{n})}{n!}|<\sum_{n=1}^{\infty}\frac{1}{n!}$.
By comparison test, we know that $\sum_{n=1}^{\infty}\frac{1}{n!}$ is convergent. So by WM test $\sum_{n=1}^{\infty}\frac{sin(x^{n})}{n!}$ converges uniformly for $x \in \mathbb R$.
Can anyone suggest me about the second part? Is the question about term by term differentiation?
Since $\sum_{n=1}^\infty \frac{\sin(x^n)}{n!}$ converges uniformly for $x\in\mathbb R$, we may differentiate term-by-term:
\begin{align} \frac{\mathsf d}{\mathsf dx} \sum_{n=1}^\infty \frac{\sin(x^n)}{n!} &= \sum_{n=1}^\infty \frac{\mathsf d}{\mathsf dx}\frac{\sin(x^n)}{n!} \\ &= \sum_{n=1}^\infty \frac{n x^{n-1}\cos(x^n)}{n!}\\ &= \sum_{n=1}^\infty \frac{x^{n-1}\cos(x^n)}{(n-1)!}\\ &= \sum_{n=0}^\infty \frac{x^n\cos(x^{n-1})}{n!}\\ \end{align} I do not know how to further simplify this, and will defer that to someone with more expertise.