Let $$A = \lim_{x\to \infty} \sum_{k=2}^\infty \sum_{p^k\le x} \frac{\ln(p)}{p^k}$$
I want to show that $\displaystyle A – \sum_{k=2}^\infty \sum_{p^k\le x} \frac{\ln(p)}{p^k} \in \mathcal{O}(\frac{1}{\sqrt{x}})$
Generally, I want to know how fast $\displaystyle \sum_{k=2}^\infty \sum_{p^k\le x} \frac{\ln(p)}{p^k}$ converges.
Some results:
It is possible to express $A$ as $\displaystyle \sum_{p \in \mathbb{P}} \frac{\ln(p)}{p(p-1)}$ and I was able to show that $\displaystyle A – \sum_{p \le x} \frac{\ln(p)}{p(p-1)} \in \mathcal{O}(\frac{1}{x})$ using Abel summation. I wasn't able to use the same technique on the double sum above. I noticed, that there is a connection to Chebyshev's functions by $\displaystyle \psi(x) – \vartheta(x) = \sum_{k=2}^\infty \sum_{p^k \le x} \ln(p)$
Also $\displaystyle A = -\sum_{k=2}^\infty P'(k)$ where $P(s)$ denotes the Prime-Zeta function, but I don't think that this is useful for the stated problem.
I tried to approximate $A$ with the sum over all natural numbers instead of primes, and approximate that with an Integral.
I tried some other minor things but nothing seems to lead anywhere. If you got some idea, any help is appreciated.
Best Answer
I'm using the Prime Number Theorem below. Let $\kappa(p,x):=\max\{1,\lfloor\ln x/\ln p\rfloor\}$; then $$R(x):=\sum_{k=2}^\infty\sum_{p^k>x}\frac{\ln p}{p^k}=\sum_{p\in\mathbb{P}}\sum_{k>\kappa(p,x)}\frac{\ln p}{p^k}=\sum_{p\in\mathbb{P}}\frac{\ln p}{p^{\kappa(p,x)}(p-1)}.$$ Note that $\kappa(p,x)=1$ if and only if $p^2>x$; otherwise $\kappa(p,x)>\frac{\ln x}{\ln p}-1$ and $p^{\kappa(p,x)}>\frac{x}{p}$.
Hence $R(x)=S(x)+T(x)$, where $$S(x)=\sum_{p^2\leqslant x}\frac{\ln p}{p^{\kappa(p,x)}(p-1)}\leqslant\frac1x\sum_{p^2\leqslant x}\frac{p\ln p}{p-1}\in\mathcal{O}(x^{-1/2})$$ because $\frac{p\ln p}{p-1}\leqslant 2\ln p\leqslant \ln x$ and $\#\{p : p^2\leqslant x\}\in\mathcal{O}\left(\frac{x^{1/2}}{\ln x}\right)$ by PNT, and $$T(x)=\sum_{p^2>x}\frac{\ln p}{p^{\kappa(p,x)}(p-1)}=\sum_{p^2>x}\frac{\ln p}{p(p-1)}\in\mathcal{O}(x^{-1/2})$$ is what you've shown already (moreover, this can't be enhanced).