I tried to make a contradiction for this by supposing this series converges on some $k\in \mathbb{R}$.
Suppose $\sum_{n=1}^{\infty }e^{-nx}\cos(nx)$ converges to some $f(x)$ pointwisely when $x=k\in \mathbb{R}$ for some $k$. Then this means, it converges to $\sum_{n=1}^{\infty }e^{-nk}\cos(nk)$
Now, we can observe that
$$\frac{1}{1-e^{k}}=-\sum \frac{1}{e^{nk}}\leq \sum \frac{\cos(nk)}{e^{nk}}\leq \sum \frac{1}{e^{nk}}=\frac{1}{e^{k}-1}$$
If $k<0$, then $\frac{1}{1-e^{k}}> 0$ and $\frac{1}{e^{k}-1}< 0$. But, there doesn't exist $f(k)$ satisfying the condition. Thus, $k>0$.
But, I can't find a method to find a contradiction for $k>0$.
Can I get some help?
Why wouldn't this converge pointwise on the set of positive reals though:
For all $x>0$ note that $$\sum_{n \in \mathbb{N}} |e^{-nx} \cos nx| \le \frac{1}{1-e^{-x}}$$ in particular is finite, which implies that the sum $\sum_{n \in \mathbb{N}} e^{-nx} cos nx$ must converge for each such $x$ as well.