Omega 2

Disclaimer: I do not have a proof that this sum converges.

$\Omega(n)$ is a completely additive function. That is, $\Omega(ab)=\Omega(a)+\Omega(b)$ , where a and b are positive integers.

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Lemma: For a completely additive function, $f*1=\frac{1}{2}f(n)\tau(n)$

Where $*$ is the Dirichlet convolution.

Proof:

By definition of completely additive,

$f(n)=f(d)+f(n/d)$

$f*1=\sum_{d\mid n} f(n/d)=\sum_{d\mid n} (f(n)-f(d))=\tau(n)f(n) - \sum_{d\mid n} f(d))=\tau(n)f(n) - f*1$

Rearranging,

$f*1=\boxed{\frac{1}{2}f(n)\tau(n)}$

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Since $\Omega$ is a completely additive function,

$2\Omega*1=\Omega(n)\tau(n)$

$2\zeta(2)\sum_{n=1}^{\infty}\frac{\Omega(n)}{n^2}=\sum_{n=1}^{\infty}\frac{\Omega(n)\tau(n)}{n^2}$

Also, since $\Omega=g*1$ , where $g(n)=\begin{cases} 1 \text{ if } n=p^a, \quad a>0 \\ 0 \text{ otherwise} \end{cases}$

$\sum_{n=1}^{\infty}\frac{\Omega(n)}{n^2}=\zeta(2)\sum_{n=1}^{\infty}\frac{g(n)}{n^2}$

$\sum_{n=1}^{\infty}\frac{g(n)}{n^2}=\sum _{p \text{ is prime}}\sum _{a=1}^{ \infty}\frac{1}{p^{2a}}=\sum _{p \text{ is prime}}\frac{1}{p^2-1}$

Putting it all together,

$\sum_{n=1}^{\infty}\frac{\Omega(n)\tau(n)}{n^2}=2\zeta(2)^2\sum _{p \text{ is prime}}\frac{1}{p^2-1}$

Using the fact that $\zeta(2)=\frac{\pi^2}{6}$ ,

$\sum_{n=1}^{\infty}\frac{\Omega(n)\tau(n)}{n^2}=\boxed{\frac{\pi^4}{18}\sum _{p \text{ is prime}}\frac{1}{p^2-1}}$