I have been tasked to show $$\boxed{\displaystyle\sum_{j=1}^{M-1} \sin\left(\frac{j\pi}{M}\right)\frac{j}{M}\left(1-\frac{j}{M}\right)=\frac{1}{4M^2}\sin\left(\frac{\pi}{M}\right)\csc^4\left(\frac{\pi}{2M}\right)}$$
I think I could do this with a mess of complex exponentials ($\displaystyle\sum_{j=1}^{M-1}\exp\left(\frac{i\pi xj}{M}\right)$ evaluated at 1) and differentiating the geometric series sum twice but that seems very messy. I was wondering if there was an easier way.
If not, is there an easy way to instead find this sum to leading order in M?
It is still manageable. Note $$f(x)= \sum_{j=1}^{M-1}e^{i\frac{\pi xj}{M}} =\frac{e^{i\frac{\pi x}{M}}-e^{i\pi x}}{1-e^{i\frac{\pi x}{M}} } $$ $$f’(x)= i \sum_{j=1}^{M-1}e^{i\frac{\pi xj}{M}}\frac{\pi j}M,\>\>\> f’’(x)= -\sum_{j=1}^{M-1}e^{i\frac{\pi xj}{M}}\frac{\pi^2j^2}{M^2} $$ Then \begin{align} &\sum_{j=1}^{M-1} \sin\frac{j\pi x}{M}\left(1-\frac{j}{M}\right)\frac{j}{M}\\ = &Im \left(\frac1{i\pi}f’(x)+\frac1{\pi^2}f’’(x)\right)\\ = &\frac{d}{dx}Im \left(\frac1{i\pi}f(x)+\frac1{\pi^2}f’(x)\right)\\ =& \frac{d}{dx}\left(\frac1{2\pi}-\frac1{2\pi M}\sin^2\frac{\pi x}2 \csc^2\frac{\pi x}{2M} \right)\\ = &\frac1{4M^2}\left(2\sin^2\frac{\pi x}2\cot\frac{\pi x}{2M}-M\sin\pi x\right) \csc^2\frac{\pi x}{2M} \end{align} Set $x=1$ to obtain \begin{align} \sum_{j=1}^{M-1} \sin\frac{j\pi}{M}\left(1-\frac{j}{M}\right)\frac{j}{M}= \frac1{2M^2}\cot\frac{\pi x}{2M} \csc^2\frac{\pi x}{2M} = \frac{1}{4M^2}\sin\frac{\pi}{M}\csc^4\frac{\pi}{2M} \end{align}