Prove That $\sum_{i=1}^{\infty}\left(\frac{1}{P_i^2}\left(1-\sum_{k=1}^{i-1}\frac{1}{P_k^2}\right)\right)=1-\frac{6}{\pi^2}$ ($P_i$ denotes the ith prime)
I recently derived this question while finding ways to express powers of natural numbers using primes.
My proof is long and tedious to work with and only works for powers of -2.
Are there similar variants to this question in history and is there a more generalised result (Im sure this is related to the Riemann Zeta Function).And if you could ,please give a better proof)
Also i checked in desmos for the first 1000 sums and it seems to be correct
Thanks a lot ☺☺☺
 A: $$\sum_{i=1}^\infty\frac1{p_i^s}\left(1-\sum_{k=1}^{i-1}\frac1{p_k^s}\right)=\frac1{2^s}+\frac1{3^s}\left(1-\frac1{2^s}\right)+\frac1{5^s}\left(1-\frac1{2^s}-\frac1{3^s}\right)=\\
\sum_{p\in P}\frac1{p^s}-\sum_{p\in P}\sum_{q\in P\setminus\{p\}}\frac1{p^sq^s}=1-\prod_{p\in P}\left(1-p^{-s}\right)=1-\frac1{\zeta(s)}, \Re(s)>1$$ by the Euler product formula.
A: This is my needlessly complicated answer without using the Euler product formula.
We can express 2+3+... as 
sum of multiples of 2 + sum of multiples of 3 excluding multiples 2 + Sum of multiples of 5 excluding multiples of 3 and 2 once + ... and so on for all primes.
Let us look at the nth term for this series.
$$T_n=P_n\left(\sum _{k=1}^{\infty }k-\left(\sum _{i=1}^{n-1}P_i\right)\left(\sum _{k=1}^{\infty }k\right)\right)$$
Observe that the same logic exists if we replace each term with 1/n^2
$$T_n=\frac{1}{P_n^2}\left(\sum _{k=1}^{\infty }\frac{1}{k^2}-\left(\sum _{i=1}^{n-1}\frac{1}{P_i^2}\right)\left(\sum _{k=1}^{\infty }\frac{1}{k^2}\right)\right)$$
$$=\frac{\pi ^2}{6}\left(\frac{1}{P_n^2}\left(1-\sum _{i=1}^{n-1}\frac{1}{P_i^2}\right)\right)$$
Therefore
$$\sum _{n=1}^{\infty }T_n=\frac{\pi ^2}{6}\left(\sum _{n=1}^{\infty }\frac{1}{P_n^2}\left(1−\sum _{i=1}^{n-1}\frac{1}{P_i^2}\right)\right)=\frac{\pi ^2}{6}-1$$
Implies
$$\begin{array}{l}\left(\sum _{n=1}^{\infty }\frac{1}{P_n^2}\left(1−\sum _{i=1}^{n-1}\frac{1}{P_i^2}\right)\right)=1-\frac{6}{\pi ^2}\end{array}$$
A: I actually think that the question is wrong. This is what I think the question is supposed to be:
Prove that $\Sigma_{i=1}^\infty \frac{1}{P_i^2} \left(1 - \Sigma_{k=1}^{i-1} \frac{1}{P_k^2} \left(1 - \Sigma_{j=1}^{k-1} \frac{1}{P_j^2}\left(...\right) \right)\right) = 1 - \frac{6}{\pi^2}$ where $P_k$ is the kth prime.
I do have a solution to this. Multiply the RHS by $\frac{\pi^2}{6}$, and the LHS by $\left( 
\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} .... \right)$, which are both the same thing.
Now the RHS is just $\frac{\pi^2}{6} - 1$, which is just $\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} ...$
I will expand the LHS and prove that the terms match that on the RHS.
Observe that the LHS is just $\left( 
\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} .... \right)\left(\Sigma \frac{1}{p_i^2} - \Sigma \frac{1}{p_i^2 p_k^2} + \Sigma \frac{1}{p_i^2 p_k^2 p_j^2} - ....\right)$ where the signs keep swapping every time and the summations are taken over all possibilities of the product of primes.
Now expand this product of infinite sums and consider the coefficient of a number $\frac{1}{n^2}$, for all $n \ge 2$, if we can prove that the coefficient is $1$, then we are done since then this will be the same as the RHS.
If $n$ is a prime, then obviously, it will appear only once in the expansion and so the coefficient is $1$.
Now, suppose $n = p_1^{\alpha_1}p_2^{\alpha_2}....p_k^{\alpha_k}$. Then, you can check that the coefficient becomes
$ 1 + \left(\binom{k}{1} - \binom{k}{2} + \binom{k}{3} ..... + (-1)^{k-1}\binom{k}{k-1}\right)$, where the second part is well known to be just $0$. So, the coefficient is always $1$ and so the proof is done
The above two solutions also work perfectly, but only when the question is changed to this. In N-N's solution, the formula for $T_n$ was incorrect as by PIE (Principle of inclusion and Exclusion), there are going to be a lot more terms as you would have overcounted and undercounted some numbers. For example, you counted the multiples of $5$ excluding mutiples of $2$ and $3$ but forgot to again add back multiples of $6$ as you had subtracted them twice.
Hope this helps :)
