Sum of reciprocal of primes failed computation Set $$X:=\sum_p \dfrac{1}{p^2}=\dfrac{1}{2^2}+\dfrac{1}{3^2} +\dfrac{1}{5^2}+\cdots$$
As $X$ is absolute convergent and less than $1$, we have (not sure for infinite rearrangement) naive calculation implies
$$\dfrac{\pi^2}{6}-1=\sum_{\substack{n\\ \Omega(n)=1}}\dfrac{1}{n^2}+\sum_{\substack{n\\ \Omega(n)=2}}\dfrac{1}{n^2}+\cdots,$$
where $\Omega(n)$ denotes the number of prime factors of $n$ counting multiple.
The right term above is
$$X+X^2+\cdots.$$
Thus, $X=1-6/\pi^2<0.4$, but with the answer in the following link this is false.
https://mathoverflow.net/questions/53443/sum-of-the-reciprocal-of-the-primes-squared
Is it possible to fill the gap of the calculation?

It's not true that $\displaystyle \sum_{\substack{n\\ \Omega(n)=2}}\dfrac{1}{n^2} = \biggl( \sum_{\substack{n\\ \Omega(n)=1}}\dfrac{1}{n^2} \biggr)^2$ (and so on). Indeed,
\begin{align*}
\biggl( \sum_{\substack{n\\ \Omega(n)=1}}\dfrac{1}{n^2} \biggr)^2 = \biggl( \sum_{p\text{ prime}}\dfrac{1}{p^2} \biggr)^2 = \sum_{p,q\text{ prime}}\dfrac{1}{p^2q^2}
\end{align*}
counts every integer that's the product of two distinct primes twice, once as $pq$ and once as $qp$. The correct (but perhaps not useful) identity is
$$
\biggl( \sum_{\substack{n\\ \Omega(n)=1}}\dfrac{1}{n^2} \biggr)^2 = 2\sum_{\substack{n\\ \Omega(n)=2}}\dfrac{1}{n^2} - \sum_{\substack{n\\ \Omega(n)=1}}\dfrac{1}{n^4}.
$$
