Does the series $f'(x) = \sum\limits_{n = 1}^{\infty} -\frac{1}{n} e^{-nx} \cos(e^{-nx})$ converge uniformly or pointwise on $\mathbb{R}$?

I would say no, since for $$x < 0$$, $$f'(x) \rightarrow \infty$$ as $$n \rightarrow \infty$$ since $$e^{-nx} \rightarrow \infty$$ as $$n \rightarrow \infty$$. So the series doesn't even converge for $$x < 0$$.

Also $$f'(0) = \sum\limits_{n=1}^{\infty} -\frac{1}{n} \cos(1)$$ which would seem to diverge to $$-\infty$$.

Correct?

• You have answered your own question. The series converges uniformly on all compact subsets of $[\delta,\infty)$, $\delta>0$. And the series converges pointwise on $(0,\infty)$. But the series diverges on $(-\infty,0]$. – Mark Viola Feb 25 at 18:23
• Additional remark: writing the $\cos$ in exponential form you can do the sum explicitly for $x\gt 0$ with the result $\frac{1}{2} \left(-\log \left(1-e^{(-1-i) x}\right)-\log \left(1-e^{(-1+i) x}\right)\right)$ This function now can be continued analytically, specifically to the region $x\lt 0$. – Dr. Wolfgang Hintze Feb 25 at 18:57
• @goblinb For a series to converge, it is necessary that the terms of the series tend to zero. For $x<0$, do the terms approach $0$? – Mark Viola Mar 13 at 17:06
• @goblinb And so, what can you conclude regarding convergence of the series of terms that do not approach zero? – Mark Viola Mar 13 at 17:31
• @goblinb You have it now. – Mark Viola Mar 13 at 17:48