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?

  • 1
    $\begingroup$ Can you do term-by-term differentiation? $\endgroup$
    – GEdgar
    Dec 13, 2019 at 1:14
  • $\begingroup$ @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. $\endgroup$
    – Dclrk
    Dec 13, 2019 at 1:23
  • $\begingroup$ 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. $\endgroup$
    – Michael
    Dec 13, 2019 at 1:36
  • $\begingroup$ @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? $\endgroup$
    – User124356
    Dec 13, 2019 at 1:44
  • $\begingroup$ Can you set $y=sin(x^n)$ so that it converges to $e^y-1$ and use the chain rule? $\endgroup$ Dec 13, 2019 at 2:27

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Michael
    Dec 14, 2019 at 19:11
  • $\begingroup$ It converges absolutely as well. What else is needed for term-by-term differentiation? $\endgroup$
    – Math1000
    Dec 14, 2019 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.