Convergence of $\sum_{n=1}^\infty \frac{\cos(nz)}{e^n}$ 
Determine where $$\sum_{n=1}^\infty \frac{\cos(nz)}{e^n}$$ is
  convergent.

I believe it should converge for $\text{Im}(z)<1$, diverge for $\text{Im}(z)>1$. My only question comes when $\text{Im}(z)=1$. Then we have $$\sum_{n=1}^\infty \frac{\cos(nz)}{e^n}= \frac12 \sum_{n=1}^\infty \frac{e^{inz}+e^{-inz}}{e^n}\\=\frac12 \sum_{n=1}^\infty \frac{e^{-n}e^{inx}+e^ne^{-inx}}{e^n}= \frac12 \sum_{n=1}^\infty \left(e^{-2n}e^{inx}+e^{-inx}\right).$$
I claim this should converge for no $x\in \mathbb{R}$, since the terms do not even go to zero in modulus. This answer, written by a more advanced student, claims it should converge for $x\neq 0$. Is he just having a slip-up, or am I missing something?
 A: Setting
$$
z=x+iy,
$$
we have
$$
\frac{\cos(nz)}{e^n}=\frac12e^{-(y+1)n}e^{inx}+\frac12e^{(y-1)n}e^{-inx}, \quad \forall z \in \mathbb{C},\quad n \in \mathbb{N}.
$$
Furthermore, the series 
$$
\sum_{n=1}^\infty e^{-(y+1)n}e^{inx}
$$
converges for every $z \in \mathbb{C}$ with $\Im z>-1$ and diverges elsewhere. 
In fact, for every $z \in \mathbb{C}$ with $\Im z \le -1$ we have
$$
\lim_n|e^{-(y+1)n}e^{inx}|=\lim_ne^{-(y+1)n}>0
$$
Similarly, the series
$$
\sum_{n=1}^\infty e^{(y-1)n}e^{-inx}
$$
converges for every $z \in \mathbb{C}$ with $\Im z<1$, and diverges elsewhere because for every $z \in \mathbb{C}$ with $\Im z\ge 1$ we have
$$
\lim_n|e^{(y-1)n}e^{-inx}|=\lim_ne^{(y-1)n} >0.
$$
Thus, the series
$$
\sum_{n=1}^\infty\frac{\cos(nz)}{e^n}
$$
converges for every $z \in \mathbb{C}$ with $-1<\Im z <1$, and diverges elsewhere. For every $z \in \mathbb{C}$ with $-1<\Im z<1$ we have
\begin{eqnarray}
\sum_{n=1}^\infty\frac{\cos(nz)}{e^n}&=&\frac12\sum_{n=1}^\infty\left[(e^{-(y+1)+ix})^n+(e^{(y-1)-ix})^n\right]=-1+\frac{1}{2(1-e^{-(y+1)}e^{ix})}+\frac{1}{2(1-e^{y-1}e^{-ix})}\\
&=&-1+\frac{e}{2(e-e^{iz})}+\frac{e}{2(e-e^{-iz})}=\frac{e^{iz}}{2(e-e^{iz})}+\frac{e^{-iz}}{2(e-e^{-iz})}
\end{eqnarray}
A: The series diverges for all $\text{Im}(z)=1$. The "proof" posted on the page you linked to only shows that the partial sums are bounded, which certainly is not equivalent to convergence.
