Question

2. Express in closed form. sum =-infinity to infinity{1/ {z^ 3 -n^ 3}}

Answer 1

$\displaystyle{\sum_{-\infin}^{\infin}}\frac{1}{z^3-n^3}=2\pi i\cdot \sum res\ f(z_0)$ for $z_0$ in the upper half-plane

1) $z_0=n$

$res_1\ f(z_0)=\displaystyle{\lim_{z\to n}}[f(z)(z-n)]=\displaystyle{\lim_{z\to n}}\frac{1}{z^2+zn+n^2}=\frac{1}{3n^2}$

2) $z_0=n(-0.5+0.5i\sqrt 3)$

$res_2\ f(z_0)=\displaystyle{\lim_{z\to z_0}}[f(z)(z-z_0)]=\displaystyle{\lim_{z\to z_0}}\frac{1}{z^2+zz_0+z_0^2}=\frac{1}{3z_0^2}=\frac{1}{3n^2(-0.5+0.5i\sqrt 3)^2}$

3) $z_0=n(-0.5-0.5i\sqrt 3)$ is not in the upper half-plane

$\displaystyle{\sum_{-\infin}^{\infin}}\frac{1}{z^3-n^3}=\frac{2\pi i}{3n^2}(1+\frac{1}{(-0.5+0.5i\sqrt 3)^2})$