I'd like to know whether the following calculations are correct, as I obtain an unexpected result. Start from $$\frac{1}{N}\sum_{n=1}^NE_n^2\left(1+\sum_{j\ne k}^N\rho_{jk}\exp\left(i\frac{2\pi n}{N}(k-j)\right)\right),$$ where $\rho$ is hermitian, and $E_n=2-2\cos(2\pi n/N)$.
The first term should give $$\frac{2}{N}\sum_n\left(1+\cos^2(2\pi n/N)-2\cos(2\pi n/N)\right)=\frac{2}{N}(N+N/2)=3.$$ As for the second, write $$\sum_{j\ne k}^N\rho_{jk}\sum_nE_n^2\exp\left(i\frac{2\pi n}{N}(k-j)\right),$$ and consider the sum $\sum_{k=1}^n(e^{i2\pi (n\cdot z)/N})^n=0$ for any integer $z$. The product $$\cos\left(\frac{2\pi n}{N}\right)\exp\left(\frac{i 2\pi n(k-j)}{N}\right)$$ can be reduced to the previous sum by applying Euler's identity on the exponential and Werner's formulae on the trigonometric products.
Likewise, the duplication formulae give $\cos^2\alpha=(1+\cos 2\alpha)/2$ and therefore the last term should sum to zero as well.
My problem is that I am expecting a result strictly greater than $4$, and I obtain $3$. Is there any mistake in my calculations? Thanks in advance.
The question has already been answered in the comments: A factor of $2$ is missing in the summation of the first term, which therefore yields $6$ and not $3$.