$\sum e^{(-nx)}\cos(nx)$ is not a uniformly convergent series?

Question

How to prove that $\sum e^{(-nx)} \cos(nx)$ is not a uniformly convergent series? Where $x$ is a positive real number.

My try: Local Minimums of $e^{(-nx)} \cos(nx)$ attain at x = $3\pi/4n +3m\pi/2n$ and min value will be a fixed negative number for each n. We can not say anything about convergence or divergence from this. Abel's test and Dirichlet's test also fail to decide anything about convergence and divergence..

Answer 1

Suppose that your series is uniformly convergent on $I=]0,+\infty[$. This means that the partial sum $S_n(x)$ converge uniformly on $I$, hence $S_{n}(x)-S_{n-1}(x)=\exp(-nx)\cos(nx)$ converge uniformly to $0$. Hence for every $\varepsilon>0$, there exist an $N$ such that for every $x\in I$ and $n\geq N$, we have $\exp(-nx)|cos(nx)|<\varepsilon$. Now take $\varepsilon<e^{-1}|\cos(1)|$, and $x=1/n$.

Answer 2

HINT: first ask yourself what domain are you considering, than use that $$ \sum_ne^{nx}\cos(nx)=\Re\left(\sum_ne^{nx(1+i)}\right) $$

Answer 3

Solution:

Step 1:

We define the partial sum $$S_m(x)=\sum_{n=0}^m e^{-nx}\cos(nx)$$

Step 2:

$\cos(nx)=1\Leftrightarrow nx=2\pi k$ for $k\in\mathbb{Z}$.

Step 3:

Consider $|S_{m}(x)-S_{m-1}(x)|=e^{-mx}|\cos(mx)|$.

Step 4:

$\sup_{x>0} |S_m(x)-S_{m-1}(x)|\geq \left|S_m\left(\frac{2\pi}{m}\right)-S_{m-1}\left(\frac{2\pi}{m}\right)\right|=e^{-2\pi}|\cos(2\pi)|=e^{-2\pi}$. Therefore $(S_m)_m$ is not convergent uniformly on $(0,\infty)$.