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 $-\infty$.


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    $\begingroup$ 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 $(-\infty,0]$. $\endgroup$ – Mark Viola Feb 25 at 18:23
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    $\begingroup$ 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 $x\lt 0$. $\endgroup$ – Dr. Wolfgang Hintze Feb 25 at 18:57
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    $\begingroup$ @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 $0$? $\endgroup$ – Mark Viola Mar 13 at 17:06
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    $\begingroup$ @goblinb And so, what can you conclude regarding convergence of the series of terms that do not approach zero? $\endgroup$ – Mark Viola Mar 13 at 17:31
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    $\begingroup$ @goblinb You have it now. $\endgroup$ – Mark Viola Mar 13 at 17:48

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