As easy as e^pi

I actually don't know the proof but

$\large{\displaystyle \sum^{\infty}_{n=1} \frac{1}{n^2-a^2}=\frac{1}{2a^2}-\frac{\pi}{2a}\cot (\pi a)}$

now $\large{\coth (iz)=\cot (z)=i \frac{e^{i z}+e^{-i z}}{e^{i z}-e^{-i z}}}$

$\large{\displaystyle \sum^{\infty}_{n=1} \frac{1}{n^2-a^2}=\frac{1}{2a^2}-\frac{i \pi}{2a}\left[ \frac { { e }^{ i\pi a }+{ e }^{ -i\pi a } }{ { e }^{ i\pi a }-{ e }^{ -i\pi a } } \right] }$

Here $a=\sqrt{2}$

$\large{\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 }-2 } } =\frac { 1 }{ 4 } -\frac { i\pi }{ 2\sqrt { 2 } } \left[ \frac { { e }^{ i\pi \sqrt { 2 } }+{ e }^{ -i\pi \sqrt { 2 } } }{ { e }^{ i\pi \sqrt { 2 } }-{ e }^{ -i\pi \sqrt { 2 } } } \right] }$

$\text{ Euler's totient function (or Euler's phi function), denoted as φ(n) or ϕ(n), is an arithmetic function that counts the positive integers less than or equal to n that are relatively prime to n}$

$\boxed{\phi(11)=10}$