To evaluate $$\sum_{d\in\mathbb Z}\sum_{c\in\mathbb Z\smallsetminus\{0\}}\left(\frac{1}{cz+d}-\frac{1}{cz+d+1}\right)$$ where $z$ on the upper half plane of $\mathbb C$.
We first consider a finite sum on $d$. $$\sum_{d=N}^{n-1}\sum_{c\in\mathbb Z\smallsetminus\{0\}}\left(\frac{1}{cz+d}-\frac{1}{cz+d+1}\right)=\sum_{c\in\mathbb Z\smallsetminus\{0\}}\sum_{d=N}^{n-1}\left(\frac{1}{cz+d}-\frac{1}{cz+d+1}\right)=\sum_{c\in\mathbb Z\smallsetminus\{0\}}\left(\frac{1}{cz-N}-\frac{1}{cz+N}\right).$$
$$=\frac{1}{z}\sum_{c\in\mathbb Z\smallsetminus\{0\}}\left(\frac{1}{c-N/z}-\frac{1}{c+N/z}\right)=\frac{1}{z}\left(\sum_{c\neq0}\frac{1}{c-N/z}+\sum_{c\neq0}\frac{1}{-c-N/z}\right)$$ We set $-N/z:=\tau$. Use the following Fourier expansion $$\frac{1}{\tau}+\sum_{c\neq0}\frac{1}{c+\tau}(=\pi\cot(\pi\tau))=\pi i-2\pi i\sum_{m>0}\exp(2\pi im\tau).$$
We can see that the only term that survives is $\pi i$ in both Fourier series when $N\to\infty$. So the result is $\color{red}{+}\frac{2\pi i}{z}$. But it should be $\color{red}{-}\frac{2\pi i}{z}$ because of this. I doublechecked serveral times but I could not find my mistake(s)..
Your expression for the Fourier series is off. It should be
\begin{align*} \frac{1}{\tau}+\sum_{c\neq 0}\frac{1}{c+\tau}&=\pi i-2\pi i\sum_{m=0}\exp(2\pi im\tau)\\ &=\pi i-2\pi i-2\pi i\sum_{m>0}\exp(2\pi i m\tau)\\ &=-\pi i-2\pi i\sum_{m>0}\exp(2\pi i m\tau), \end{align*} which hopefully fixes your error. I've seen some authors start this sum at $m=0$ and others at $m=1$ so there is definitely cause for confusion.