Prove Convergence of function series $\sin(nx)e^{-nx}$. I'm having trouble with this. I need to prove the convergence and the continuity of the convergent function for the function:
$f_n(x) = e^{-nx}\sin(nx)$
And I'm having trouble. I showed that it's less than $\dfrac{e^{-x}}{e-1}$, but... that doesn't help me!
What can I do here to move forward?
Thanks!
 A: Pointwise, for any $x\in\mathbb{R}^+$ we have:
$$ \lim_{n\to +\infty} e^{-nx}\sin(nx) = 0$$
since for any real number $y$ we have $|\sin y|\leq 1$ and for any positive number $n$ we have $\lim_{n\to +\infty}e^{-nx}=0$. The convergence cannot be uniform, since for every $n\in\mathbb{N}$ we have:
$$ f_n\left(\frac{1}{n}\right) = \frac{\sin 1}{e}>0. $$
Additionally every $f_n$ belongs to $L^1(\mathbb{R}^+)\cap L^2(\mathbb{R}^+)$ and:
$$ \int_{0}^{+\infty} e^{-nx}\sin(nx)\,dx = \frac{1}{n}\int_{0}^{+\infty}\frac{\sin z}{e^z}\,dx = \frac{1}{2n},$$
$$ \int_{0}^{+\infty} e^{-2nx}\sin^2(nx)\,dx = \frac{1}{n}\int_{0}^{+\infty}\frac{\sin^2 z}{e^{2z}}\,dz = \frac{1}{8n},$$
so we have convergence to zero with respect to both the $L^1$ and the $L^2$ norm. Now, the condition $|f_n(x)|\leq e^{-nx}$ is enough to grant the the series:
$$ g(x) = \sum_{n\geq 1}f_n(x) = \sum_{n\geq 1}e^{-nx}\sin(nx) = \text{Im}\sum_{n\geq 1}e^{-n(x-i)} $$
is convergent for any $x>0$. Since the last sum is a geometric series for any $x>0$ we also have:
$$ g(x) = \text{Im}\left(\frac{e^i}{e^x-e^i}\right)=\color{red}{\frac{e^x \sin 1}{1+e^{2x}-2e^x\cos 1}}. $$
Here $\text{Im}$ stands for the imaginary part of a complex number, $\text{Im}(a+ib)=b$. 
$\text{Im}$ commutes with $\sum$ since it is a linear operator.
