I came across this claim about this series on prime number powers: $$ \sum_{p \in\mathbb P} \sum_{k \in \mathbb N} \frac{1}{k}\frac{\chi(p^k)}{p^k} = \ln\left(\frac\pi4\right)$$ where $\mathbb P$ is the set of primes, $\chi(n)$ is defined to be $0$ if the input is even, $-1$ if $n$ is congruent to $3$ modulo $4$, and $1$ if $n$ is congruent to $1$. I didn't see any proof of this, but taking the sum for all $p^k \leq 1,000,000$ matches the first four digits, so it seems likely to me.
This was said in relation to the famous series $ \sum_{n \in \mathbb N} \frac{\chi(n)}{n} = \frac\pi4$, so I assume there is some relationship between the transformation on the left-hand side and taking the logarithm on the right hand side.
Is there anything to this connection? Or any other arguments for why this series converges to the given value? Or does it just happen to be very close and I got pranked?
Not a full proof, but it will show the relationship.
$\chi(n)$ is a completely multiplicative function ($\chi(ab)=\chi(a)\chi(b),)$ so if you define $$f(s)=\sum_{n=1}^{\infty} \frac{\chi(n)}{n^s},$$ $f$ can be factored:
$$f(s)=\prod_p \sum_{k=0}^{\infty}\left(\frac{\chi(p)}{p^{s}} \right)^k=\prod_{p} \frac{1}{1-\chi(p)p^{-s}}$$
So $$\ln f(s)=-\sum_{p} \log(1-\chi(p)p^{-s})$$
Then use the power series for $\log(1-\chi(p)p^{-s})$ when $s=1.$
There might be some difficulty with the series transformation, because the sums when $s=1$ do not converge absolutely. But this is the fundamental source of the relationship. It all works when $\operatorname{Re}s>1,$ since then everything converges absolutely.
So $$\ln(f(s))=\sum_p\sum_{k=1}^{\infty}\frac{\chi(p^k)}{kp^{sk}}$$ when $\operatorname{Re}s>1,$ but I'm unsure what you need for $s=1.$