It's not obvious for me that $$ \sum_{r=-n}^n (n+1-|r|)\exp(irs) = \left(\sum_{k=0}^n \exp(i(k-n/2)s)\right)^2 $$
I checked with Mathematica the correctness of the formula. Then, I tried :
- Cauchy-Schwarz,
- separate positive indices and nonpositive indices in the lhs,
- change the indice $k = r+n/2$,
- Write
$$ \sum_{r=-n}^n (n+1-|r|)\exp(irs) = \sum_{j=0}^n \sum_{r=-j}^j \exp(irs), $$
Start with the RHS: $$ \left(\sum_{k=0}^{n} \exp(is(k-n/2))\right)^2=\left(\exp(-isn/2)\sum_{k=0}^{n} \exp(isk)\right)^2 $$This is a geometric series: $$ =\exp(-isn)\cdot \left(\frac{1-\exp(is(n+1))}{1-\exp(is)}\right)^2 $$ $$ =\exp(-isn)\cdot \frac{1-2\exp(is(n+1))+\exp(2is(n+1))}{1-2\exp(is)+\exp(2 i s)} $$ $$ = \frac{\exp(-i s n)-2\exp(is)+\exp(is(n+2))}{1-2\exp(is)+\exp(2 i s)} $$ $$ = \frac{\exp(-i s (n+1))-2+\exp(is(n+1))}{\exp(-is)(1-2\exp(is)+\exp(2 i s))} $$ $$ = \frac{-2+2\cos(s(n+1))}{-2+2\cos(s)} $$ $$ =\frac{1-\cos((n+1)s)}{1-\cos(s)} $$Now on the LHS we have: $$ (n+1)\sum_{r=-n}^n \exp(i r s) =(n+1)\sum_{r=-n}^{n} \cos(r s) $$ $$ =2(n+1)\left(-1+\sum_{r=0}^{n}\cos(r s)\right) $$This sum is well-known, and we can use angle-addition to clean things up: $$ =2(n+1)\left(-1+\csc \left(\frac{s}{2}\right) \sin \left(\frac{1}{2} (n+1) s\right) \cos \left(\frac{n s}{2}\right)\right) $$ $$ =(n+1)\csc \left(\frac{s}{2}\right) \sin \left(\frac{1}{2} (n+1) s\right) $$Lastly, we have $$ -\sum_{r=-n}^{n}|r| \exp(i r s)=-2\sum_{r=0}^{n} r\cos( r s) $$This sum also admits a closed-form, though not as nice as the other two: $$ =-2\cdot \frac{1-n \cos (n s)-\cos (n s)+n \cos ((n+1) s)}{1-\cos (s)} $$ $$ =\frac{1-(n+1) \cos (n s)+n \cos ((n+1) s)}{1-\cos(s)} $$Gluing the LHS together and slogging through the angle addition does the trick. $$ LHS (1-\cos(s)) $$ $$ = (n+1)\csc \left(\frac{s}{2}\right) \sin \left(\frac{1}{2} (n+1) s\right)(1-\cos(s))+(1-(n+1) \cos (n s)+n \cos ((n+1) s)) $$ $$ =1-\cos((n+1) s)=RHS (1-\cos(s)) $$