<s>[INST] How did they simplify this sum? Im having trouble seeing how this simplification is justified:
$$\frac{2}{\pi}\sum_{k=-\infty}^\infty \frac{e^{i(2k+1)x}}{(2k+1)^{2}} =\frac{4}{\pi}\sum_{k=0}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}} $$
There are no steps on how they simplified, could anyone help me with it?
 [/INST] Note that we can write
$$\frac{2}{\pi}\sum_{k=-\infty}^\infty \frac{e^{i(2k+1)x}}{(2k+1)^{2}} =\frac{2}{\pi}\sum_{k=-\infty}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}+i\frac{2}{\pi}\sum_{k=-\infty}^\infty \frac{\sin(2k+1)x}{(2k+1)^{2}} \tag1$$

Next, we split the series for the cosine into series over negative and non-negative values of the index to find 
$$\begin{align}
\sum_{k=-\infty}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}&=\sum_{k=-\infty}^{-1} \frac{\cos(2k+1)x}{(2k+1)^{2}}+\sum_{k=0}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}\\\\
&=\sum_{k=1}^\infty \frac{\cos(2k-1)x}{(2k-1)^2}+\sum_{k=0}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}\\\\
&=\sum_{k=0}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}+\sum_{k=0}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}\\\\
&=2\sum_{k=0}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}\tag 2
\end{align}$$

Similarly, we split the series for the sine into series over negative and non-negative values of the index to find 
$$\begin{align}
\sum_{k=-\infty}^\infty \frac{\sin(2k+1)x}{(2k+1)^{2}}&=\sum_{k=-\infty}^{-1} \frac{\sin(2k+1)x}{(2k+1)^{2}}+\sum_{k=0}^\infty \frac{\sin(2k+1)x}{(2k+1)^{2}}\\\\
&=-\sum_{k=1}^\infty \frac{\sin(2k-1)x}{(2k-1)^2}+\sum_{k=0}^\infty \frac{\sin(2k+1)x}{(2k+1)^{2}}\\\\
&=-\sum_{k=0}^\infty \frac{\sin(2k+1)x}{(2k+1)^{2}}+\sum_{k=0}^\infty \frac{\sin(2k+1)x}{(2k+1)^{2}}\\\\
&=0\tag 3
\end{align}$$

Putting together $(1)-(3)$ yields the coveted result.
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