# How fast does this sum over primes converge?

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.

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).
• +1 and a big thank you for writing $\in \mathcal{O}(\cdot)$ instead of $= \mathcal{O}(\cdot)$ – user3733558 Mar 24 at 9:38
• @user3733558 people write $= O(\cdot)$ to mean $\in O(\cdot)$, cause it's much easier that way (e.g., you can write $f = g+O(h)$ to mean $f-g \in O(h)$). – mathworker21 Mar 24 at 20:24