Study the convergence of this series

Question

Study the uniform convergence of the following series:

$$\sum_{n=1}^\infty e^{-nx} \sin(nx), \qquad x\in[1,\infty)$$

$$\sum_{n=1}^\infty e^{-nx} \sin(nx)\leq \sum_{n=1}^\infty e^{-nx}$$

We know that $n^2e^{-nx} \longrightarrow 0$ so $\sum_{n=1}^\infty e^{-nx} \leq \sum_{n=1}^\infty \frac{1}{n^2}$ and so $\sum_{n=1}^\infty e^{-nx}$ converges as well as the series under study.

For uniform convergence, I can't do any step!

I found this question in a book without answers, and I'm familiar with proving uniform convergence using normal convergence or using the remainder term ( $R_n$ converges uniformly to $0$), not any other way.

I tried to find $f'_n(x)$ in order to find its maximum but got that $f'_n(x)=0$ for $x= \frac{\pi}{4n} + \frac{\pi k}{n}$. Is this a point for maximum or minimum? And if it's for maximum, how to find this maximum, as it may be positive or negative depending on our $k$. So is this normally convergent?

I really appreciate your help!!

Answer 1

The uniform convergence for $x\in[1,\infty)$ is evident since we have

$$\left|\sum_{n=1}^\infty e^{-nx}\sin(nx)\right|\le \sum_{n=1}^\infty e^{-nx}\le \sum_{n=1}^\infty e^{-n}<\infty$$

The Weierstrass M-Test guarantees that the series $\sum_{n=1}^\infty e^{-nx}\sin(nx)$ converges uniformly on $[1,\infty)$.

---

Alternatively, we see that for any $\epsilon>0$ we have

$$\left|\sum_{n=N}^\infty e^{-nx}\sin(nx)\right|\le \sum_{n=N}^\infty e^{-nx}\le \sum_{n=N}^\infty e^{-n}<\frac{e^{-N}}{1-e^{-1}}<e^{-N}<\epsilon$$

whenever $N>\max\left(1,-\log(\epsilon)\right)$.

Answer 2

Hint : To prove uniform convergence you need to argue the remainder of the series $\sum_{n=N}^\infty e^{-nx}\sin(nx)$ is bounded above on $[1, +\infty[$ by something that goes to $0$ as $k$ goes to $+\infty$.

Therefore compute $\sum_{n=N}^\infty e^{-nx}\sin(nx) = Im(\sum_{n=N}^\infty e^{-nx+inx})$ and see if it is bounded above by something that goes to $0.$

Answer 3

Potentially useful:

$$ e^{-nx}\sin(nx)=e^{-nx}\frac{(e^{inx}-e^{-inx})}{2i}=\frac{1}{2i}((e^{ix-x})^n-(e^{-ix-x})^n) $$

$$ \sum_{n=1}^{\infty}e^{-nx}\sin(nx)=\frac{1}{2i}\lbrace\sum_{n=1}^{\infty}(e^{ix-x})^n-\sum_{n=1}^{\infty}(e^{-ix-x})^n\rbrace $$

$$ \sum_{n=1}^{\infty}e^{-nx}\sin(nx)=\frac{1}{2i}\lbrace\frac{e^{ix-x}}{1-e^{ix-x}}-\frac{e^{-ix-x}}{1-e^{-ix-x}}\rbrace $$

$$ \frac{e^{ix-x}}{1-e^{ix-x}}-\frac{e^{-ix-x}}{1-e^{-ix-x}}=- \frac{\sin(x)i}{\cos(x)-\cosh(x)} $$

$$ \sum_{n=1}^{\infty}e^{-nx}\sin(nx)=\frac{1}{2}\frac{\sin(x)}{\cosh(x)-\cos(x)} $$