OK, bear with me the whole way through because this gets interesting (sorry about the set up, I don't know how to use the maths thing)

I found an alternate expression for the sum of the reciprocals of the squares of the primes using a simple sieving technique:

\sum_{n\geq 1} \frac{1}{n^2} - 1 - \left(\sum_{n\geq 1} \frac{1}{n^2} -1\right)\left(\frac{1}{2^2} + \frac{1}{3^2}\left(1-\frac{1}{2^2}\right) + \frac{1}{5^2}\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \frac{1}{7^2}\left(1 - \frac{1}{5^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \ldots\right) =

= \frac{\pi^2}{6} - 1 - \left(\frac{\pi^2}{6} - 1\right)\left(\frac{1}{2^2} + \frac{1}{3^2}\left(1 - \frac{1}{2^2}\right) + \frac{1}{5^2}\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \frac{1}{7^2}\left(1 - \frac{1}{5^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \ldots\right)

Now I managed to prove that this equals pi

^{2}/20. As such I found that the sieve:

\left(\frac{1}{2^2} + \frac{1}{3^2}\left(1 - \frac{1}{2^2}\right) + \frac{1}{5^2}\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \frac{1}{7^2}\left(1 - \frac{1}{5^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \ldots\right) = \frac{7\pi^2 - 60}{10\pi^2 - 60}

Now, using the same sieve for the sum of the reciprocals of the primes we get:

\sum_{n\geq 1} \frac1n - 1 - \left(\sum_{n\geq 1} \frac1n -1\right)\left(\frac12 + \frac13 \left(1 - \frac12\right) + \frac15 \left(1 - \frac13\right)\left(1 - \frac12\right) + \frac17 \left(1 - \frac15\right)\left(1 - \frac13\right)\left(1 - \frac12\right) +\ldots\right) = \sum_{p} \frac1p

Now any mathematician worth their salt knows that both the harmonic series and the reciprocal prime series converge, but the above sieve section does not because some rearranging gives:

1 - \frac{\sum_{p} \frac1p }{\sum_{n \geq 1} \frac1n - 1} = \frac12 + \frac13 \left(1 - \frac12\right) + \frac15 \left(1 - \frac13\right)\left(1 - \frac12\right) + \frac17 \left(1 - \frac15\right)\left(1 - \frac13\right)\left(1 - \frac12\right) + \ldots

And:

\sum_{p} \frac1p < \sum_{n\geq 1} \frac1n - 1

So my question is this, what is the closed form of

\frac12 + \frac13 \left(1 - \frac12\right) + \frac15 \left(1 - \frac13\right)\left(1 - \frac12\right) + \frac17 \left(1 - \frac15\right)\left(1 - \frac13\right)\left(1 - \frac12\right) + \ldots

Any ideas at all would be helpful, it's a hard problem and I need some inspiration.

If someone can type my equations using the maths thing I'll edit those into this post