How did that pi get there?

I don't have a proof that the sum converges, so maybe somebody can fill up on that. So for my solution, I will be assuming that the sum converges.

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Let primes $p_{1,k}$ and $p_{-1,k}$ be primes that satisfy the following:

$\operatorname{mod}\left(p_{1,k},4\right)=1\\\operatorname{mod}\left(p_{-1,k},4\right)=-1$

Where $p_{n,1}<p_{n,2}<p_{n,3}...$

Let the set that contains all primes $p_{n,k}$ be $S_{n}$ . It is known that every prime except 2 and 3 belong to sets $S_{1}$ and $S_{-1}$ .

First thing to notice is that for a square-free number to be of the form $4n-3$ , it has to be a product of an even number of primes from set $S_{-1}$ and any number of primes from set $S_{1}$ .

So, the sum can be re-written to be:

$\sum_{n=1}^{\infty} \dfrac{\mu(4n-3)}{4n-3}=1-\sum \frac { 1 }{ p_{ 1 } } +\left( \sum _{ k>j } \frac { 1 }{ p_{ 1,k }p_{ 1,j } } +\sum _{ k>j } \frac { 1 }{ p_{ -1,k }p_{ -1,j } } \right) -\left( \sum _{ a>b>c } \frac { 1 }{ p_{ 1,a }p_{ 1,b }p_{ 1,c } } +\sum _{ a>b } \frac { 1 }{ p_{ -1,a }p_{ -1,b } } \sum \frac { 1 }{ p_{ 1 } } \right) +\left( \sum _{ a>b>c>d } \frac { 1 }{ p_{ 1,a }p_{ 1,b }p_{ 1,c }p_{ 1,d } } +\sum _{ k>j } \frac { 1 }{ p_{ 1,k }p_{ 1,j } } \sum _{ k>j } \frac { 1 }{ p_{ -1,k }p_{ -1,j } } +\sum _{ a>b>c>d } \frac { 1 }{ p_{ -1,a }p_{ -1,b }p_{ -1,c }p_{ -1,d } } \right) ...$ $=\frac{1}{2}\left(\prod _{p_1}^{ }\left(1-p_1^{-1}\right)\prod _{p_{-1}}^{ }\left(1+p_{-1}^{-1}\right)+\prod _{p_1}^{ }\left(1-p_1^{-1}\right)\prod _{p_{-1}}^{ }\left(1-p_{-1}^{-1}\right)\right)$ $=\frac{1}{2}\left(\prod _{p_1}^{ }\left(1-p_1^{-1}\right)\prod _{p_{-1}}^{ }\left(1+p_{-1}^{-1}\right)+\left[\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\right]^{-1}\prod _{p \text{ is prime}}^{ }\left(1-p^{-1}\right)\right)$

We will first evaluate $\prod _{p \text{ is prime}}^{ }\left(1-p^{-1}\right)$ . Through Euler Product

$\prod _{p \text{ is prime}}^{ }\left(1-p^{-s}\right)=\frac{1}{\zeta{(s)}}$

Where $\zeta{(s)}$ is the Riemann Zeta Function. Since $\zeta{(1)}$ diverges

$\prod _{p \text{ is prime}}^{ }\left(1-p^{-1}\right)=0$

Now we will focus on prime product

$\prod _{p_1}^{ }\left(1-p_1^{-1}\right)\prod _{p_{-1}}^{ }\left(1+p_{-1}^{-1}\right)=\prod _{p \text{ is prime}}^{ }\left(1-\chi(p)p^{-1}\right)$

Where $\chi(n)$ is the Dirichlet character,

$\chi (n)=\begin{cases} 1 \text{ if } \mod{(n,4)}=1 \\ -1 \text{ if } \mod{(n,4)}=-1\\ 0 \text{ otherwise} \end{cases}$

Since $\chi{(n)}$ is completely multiplicative, through Euler Product: (This prime product is already known, and it is known to be convergent)

$\prod _{p \text{ is prime}}^{ }\left(1-\chi(p)p^{-1}\right)=\left[{\sum _{n=1}^{\infty}\frac{\chi(n) }{n}}\right]^{-1}$

$\sum _{n=1}^{\infty}\frac{\chi(n) }{n}=\sum _{n=0}^{\infty}\frac{\left(-1\right)^n}{2n+1}=\arctan(1)=\frac{\pi}{4}$

Putting it all together,

$\sum_{n=1}^{\infty} \dfrac{\mu(4n-3)}{4n-3}=\frac{1}{2}\left(\frac{4}{\pi }+\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\cdot 0\right)=\frac{2}{\pi }$