Consider the limit $\lim_{n \rightarrow \infty} \sum_{k=0}^n \frac{\Lambda(4k+3)}{(4k+3)}-\frac{1}{2}\ln(4n+3)$
Where $\Lambda(n)$ is the vonmangoldt function, that is equal to zero if n is not a prime power, and equal to $\ln(p)$ if $n=p^j$ with p prime. Now from that definition I can re-write
$\lim_{n \rightarrow \infty}\sum_{k=0}^n\frac{\Lambda(4k+3)}{(4k+3)}-\frac{1}{2}\ln(4n+3)=\lim_{n \rightarrow \infty} \sum_{p^j\equiv3\,\mathrm{mod}\,4, p^j\leq n} \frac{\ln(p)}{p^{j}}-\frac{1}{2}\ln(n)$ But $p^{2n}\equiv3\,\mathrm{mod}\, 4$, has no solutions for any integer $n$ and any prime $p$,
So I need only deal with the prime powers of odd index $p^{2n+1}$,
But sence $p^{2n+1}\equiv3\,\mathrm{mod}\,4$ , always has the solution $p=3$ , which corolates with the first prime $p$ congruent to 3 modulo 4 , and I can keep adding multiples of four to p, so thats p is still prime and $p^{2n+1}\equiv3\,\mathrm{mod}\,4$,
Doesn't that mean I can re-write $\lim_{n \rightarrow \infty} \sum_{p^j\equiv3\,\mathrm{mod}\,4, p^j\leq n} \frac{\ln(p)}{p^{j}}-\frac{1}{2}\ln(n)=\lim_{n \rightarrow \infty} \sum_{p\equiv3\,\mathrm{mod}\,4, p\leq n} \ln(p)\left(\frac{1}{p}+\frac{1}{p^3}+\frac{1}{p^5}+\cdots\right)-\frac{1}{2}\ln(n)$ Which is equal to $ \lim_{n \rightarrow \infty} \sum_{p\equiv3\,\mathrm{mod}\,4, p\leq n} \frac{\ln(p)p}{p^2-1}-\frac{1}{2}\ln(n)$
Or is my reasoning wrong?, can I not re-write $\displaystyle \lim_{n \rightarrow \infty} \sum_{k=0}^n \frac{\Lambda(4k+3)}{(4k+3)}-\frac{1}{2}\ln(4n+3)$, as the series above?