<s>[INST] 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?
 [/INST] $X^2$ and $\sum_{n, \Omega(n) = 2} \frac{1}{n^2}$ are distinct: the second one only counts each $\frac{1}{p^2q^2}$ once, while $X^2$ will count both the pair $\frac{1}{p^2q^2}$ and $\frac{1}{q^2p^2}$, thanks to distributivity, which adds up to $\frac{2}{p^2q^2}$.
The same logic works for all the higher $k$, the number of times each $\frac{1}{p_1^2 \dots p_k^2}$ will appear in $X^k$ is $k!$, meaning:
$$X^k \geq k! \sum_{n, \Omega(n) = k} \frac{1}{n^2}$$
Why the $\geq$? Because $X^k$ will also contain numbers of the form $\frac{1}{p^{2k}}$, $\frac{1} {p^{2(k-1)}q^2} $ and so on! Meaning it's going to be a more complicated expression sadly.
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