Differentiability of series

Question

Is series $$\sum_{n=0}^\infty e^{-nx} \cos{nx}$$ differentiable on $(0,\infty)$?

I had this question on my exam and didn't solve it.

Any help is welcome. Thanks in advance!

Answer 1

$\sum \cos (nx)e^{-nx}$ is uniformly convergent in $[a,b]$ if $0<a<b<\infty$. So is the derived series $\sum (-n\sin (nx) e^{-nx} -n\cos (nx) e^{-nx})$ because it it dominated by $\sum 2ne^{-na}$ which converges by ratio test. These two fact are enough to conclude that $\sum \cos (nx)e^{-nx}$ is differentiable on $[a,b]$ and its derivative is $-\sum n\sin (nx) e^{-nx}$ if $0<a<b<\infty$. Since $a$ and $b$ are arbitrary we are done.