Finding $\sum_{k=1}^{n-1} \frac{1}{1-e^{2i\pi k/n}}$

Question

We know that $$\frac{z^n-1}{z-1}=(z-z_1)(z-z_2)(z-z_3).....(z-z_{n-1}), z_k=e^{2i\pi k/n}$$ We get the well known result $$(1-z_1)(1-z_2)(1-z_3).....(1-z_{n-1})=\lim_{z \to 1} \frac{z^n-1}{z-1}=n.$$ Next, we have $$\ln \left(\frac{z^n-1}{z-1}\right)=\ln(z-z_1)+\ln(z-z_2)+\ln(z-z_3)+...+\ln(z-z_{n-1})$$ D.w.r.t. $z$, we get $$\frac{z-1}{z^n-1} \frac{(z-1)nz^{n-1}-(z^n-1)}{(z-1)^2}=\frac{1}{z-z_1}+\frac{1}{z-z_2}+\frac{1}{z-z_3}+....+\frac{1}{z-z_{n-1}}$$ $$\implies F= \frac{1}{1-z_1}+\frac{1}{1-z_2}+\frac{1}{z-z_3}+....+\frac{1}{1-z_{n-1}}=\lim_{z\to 1} \frac{nz^n-nz^{n-1}-z^n+1}{z^{n+1}-z-z^n+1} \to \frac{0}{0}.$$ By L'hospital, we get $$F=\lim_{z\to 1}\frac{n^2z^{n-1} -n(n-1)z^{n-2}-nz^{n-1}}{(n+1)z^n-1-nz^{n-1}}\to \frac{0}{0}.$$ L'Hospital, once more gives $$F=\lim_{z\to 1} \frac{n^2(n-1)z^{n-2}-n(n-1)(n-2)z^{n-3}-n(n-1)z^{n-2}}{n(n+1)z^{n-1}-n(n-1)z^{n-2}}\to \frac{n-1}{2}.$$ The question is how else we can get this result?

Answer 1

Since this problem already received nice solutions in the past, consider the continuous case $$I_n=\int_1^{n-1}\frac{dk}{1-e^{\frac{2 i \pi k}{n}}}=\frac {n- 2}2$$