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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?

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No; in the equation after "Doesn't that mean I can re-write", you included lots of terms in the sum on the right-hand side that aren't included on the left-hand side: On the left-hand side, the cutoff is $p^j\le n$, whereas on the right-hand side it's $p\le n$, which comes with all terms with $p^j$, including with $p^j\gt n$. It's not clear why this shouldn't change the limit.

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I doesn't change the limit because ln(p)/p^k summed over the primes converges for all k greater then 1, and those terms would eventually appear after large enough n –  Ethan Dec 9 '12 at 22:21

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