Finding the positive integer numbers to get $\frac{\pi ^2}{9}$ As we know, there are many formulas of $\pi$ , one of them $$\frac{\pi ^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}......
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and this $$\frac{\pi ^2}{8}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}......$$
Now,find the positive integer numbers $(a_{0}, a_{1}, a_{2}....)$ to get $$\frac{\pi^2 }{9}=\frac{1}{a_{0}^2}+\frac{1}{a_{1}^2}+\frac{1}{a_{2}^2}....$$

This is equal to
$$\frac1{72} \sum_{n=-\infty}^{\infty} \left [\frac1{(n+1/6)^2} + \frac1{(n+5/6)^2}\right ] $$
This may be treated using the following formula from residue theory:
$$\sum_{n=-\infty}^{\infty} f(n) = -\pi \sum_k \operatorname*{Res}_{z=z_k} f(z) \cot{\pi z} $$
where $z_k$ are the poles of $f$.  In this case
$$f(z) = \frac1{(z+1/6)^2} + \frac1{(z+5/6)^2} $$
which has double poles at $z=-1/6$ and $z=-5/6$.  Then the sum is equal to
$$\pi \left [-\frac{\cot{(\pi/6)}}{(4/6)^2} + \pi \csc^2{(\pi/6)} -\frac{\cot{(5\pi/6)}}{(-4/6)^2} + \pi \csc^2{(5\pi/6)}\right ]  = 8 \pi^2$$
The result follows.
