uniform convergence of the series $\sum_{n =1}^{\infty}e^{-nx} sin (nx) $

Question

Prove that the series $\sum_{n =1}^{\infty}e^{-nx} sin (nx) $ for $x > 0$ is uniformly convergent.

Idea: It satisfies the neccessary condition as $a_n \rightarrow 0$.

Now use M-test for series $$|e^{-nx} sin (nx)| \le e^{-nx} , x >0$$

but $M_n = e^{-nx}$ is a geometric series convergent for for $x >0$.

Hence the given series is convergent by M-test.

Is the conclusion and the proof correct?

Answer 1

The series converges uniformly for $x \in [a, \infty)$ with $a > 0$, since

$$|e^{-nx} \sin(nx)| \leqslant e^{-na},$$

and the M-test applies.

The convergence is not uniform on $(0, \infty)$ since $f_n(x) = e^{-nx} \sin(nx)$ does not converge to $0$ uniformly. Take a sequence $x_n = 1/n$ to find $f_n(x_n) = e^{-1} \sin(1) \neq 0$.