# Show that $\sum_{n=1}^{\infty}\frac{sin(x^{n})}{n!}$ converges uniformly for $x \in \mathbb R$.

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?

• Can you do term-by-term differentiation? Dec 13, 2019 at 1:14
• @User124356 Recall that $e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} \ldots$ so then your infinite sum $\sum_{n=1}^{\infty}\frac{1}{n!} = e-1$ . I don't know if this will help I just saw that and thought I would add this. Dec 13, 2019 at 1:23
• Your proof of uniform convergence looks weird. Instead of bounding the sum, you should bound each term. If WM stands for Weirsrass M test, then you indeed need to bound each term. Dec 13, 2019 at 1:36
• @Michael do you mean by this $\sum_{n=1}^{\infty}\frac{sin(x^{n})}{n!}=\frac{sin(x)}{1!}+\frac{sin(x^{2})}{2!}+\frac{sin(x^{3})}{3!}+.........+\frac{sin(x^{n})}{n!}$. Then I take mod to bound the terms? Dec 13, 2019 at 1:44
• Can you set $y=sin(x^n)$ so that it converges to $e^y-1$ and use the chain rule? Dec 13, 2019 at 2:27

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.
• I think the assertion is incomplete, for example: Define $f_n(x) =\frac{\sin\left(x^{(n^4)}\right)}{n^2}$ for $n \in \{1, 2, 3, …\}$. Then $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly for $x \in \mathbb{R}$, but we cannot differentiate term-by-term. Dec 14, 2019 at 19:11