does this trigonometric series vanish? I have a question regarding a trigonometric series:

Shouldn't

$$\sum_{n,m}^\infty s^{n+m}sin\left((n-m)\phi\right)=0$$

where $s \in \mathbb{R}$ because the sinusoidal function is antisymmetric in $n$ and $m$ and vanishes for $n=m$?
$$\begin{align}
&amp; &amp;\sum_{n,m}^\infty s^{n+m}\sin\left((n-m)\phi\right)\\&amp;=&amp;
\text{Im}\sum_{n,m}^\infty s^{n+m}\text{e}^{(n-m)\text{i}\phi}\\&amp;=&amp;
\text{Im}\sum_{n}^\infty (s\text{e}^{\text{i}\phi})^n \sum_{m}^\infty (s\text{e}^{-\text{i}\phi})^m\\&amp;=&amp;
\text{Im}\frac{1}{1-s\text{e}^{\text{i}\phi}}\cdot\frac{1}{1-s\text{e}^{-\text{i}\phi}}\\
&amp;=&amp;\text{Im}\frac{1}{1-2\cos\phi s +s^2}\\
&amp;=&amp;0.
\end{align}
$$
Well,

\begin{align}
\sum_{n,m}^\infty s^{n+m}\sin\left((n-m)\phi\right)&amp;=\\
\sum_{n&lt;m}^\infty s^{n+m}\sin\left((n-m)\phi\right)+\sum_{n&gt;m}^\infty s^{n+m}\sin\left((n-m)\phi\right)+\sum_{n=m}^\infty s^{n+m}\sin\left((n-m)\phi\right)&amp;=\\
\sum_{n&lt;m}^\infty s^{n+m}\sin\left((n-m)\phi\right)+\sum_{n&lt;m}^\infty s^{m+n}\sin\left((m-n)\phi\right)+\sum_{n=m}^\infty s^{n+m}\sin\left(0\cdot\phi\right)&amp;=\\
\sum_{n&lt;m}^\infty s^{n+m}\sin\left((n-m)\phi\right)-\sum_{n&lt;m}^\infty s^{m+n}\sin\left((n-m)\phi\right)+0&amp;=\\
0&amp;
\end{align}