Prove that $\sum_{p\leq x}\frac{1}{p}=\log \log x+O(1)$ using PNT

Question

My problem is about proving $\sum_{p\leq x}\frac{1}{p}=\log \log x+O(1)$. The only hint I got was Prime Number Theorem. First, I used the partial summation to rewrite the sum, namely $$\sum_{p\leq x}\frac{1}{p}=\sum_{p\leq x}\frac{\log p}{p}\frac{1}{\log x}+\int_{2}^{t}\sum_{p\leq t}\frac{\log p}{p} \frac{1}{t\log^2 t} \,\mathrm{d}t.$$ But in doing so, I also need to prove that $\sum_{p\leq x}\frac{\log p}{p}=\log x+O(1)$ as a lemma. I have been working with this lemma for hours, and can't reach the conclusion. Using the hint, we have $\psi(x)/x=1 + o(1)$. From this, we conclude that $\psi(x)=O(x)$. I have proven that $\log n=\sum_{d\mid n}\Lambda (n)$, and so $$\sum_{n\leq x}\log n=x\sum_{d\leq x}\frac{\Lambda (d)}{d}+O(\psi(x))=x\sum_{d\leq x}\frac{\Lambda (d)}{d}+O(x).$$ But $\sum_{n\leq x}\log n=x\log x - x +O(\log x)$. Comparing those two, we have $$\sum_{d\leq x}\frac{\Lambda(d)}{d}=\log x + O(1)$$ How do I then use this to prove the lemma? If this seems too much, is there then a better and shorter way to answer the problem with a given hint?

Answer 1

Hint: note that $$\sum_{n\leq N}\frac{\Lambda\left(n\right)}{n}-\sum_{p\leq N}\frac{\log\left(p\right)}{p}\leq\sum_{p\leq N}\log\left(p\right)\left(\frac{1}{p^{2}}+\frac{1}{p^{3}}+\dots\right) $$ $$=\sum_{p\leq N}\frac{\log\left(p\right)}{p\left(p-1\right)}\leq\sum_{n\geq2}\frac{\log\left(n\right)}{n\left(n-1\right)}$$ and the last series is convergent. The Daniel Fischer's way is surely shorter. Using the PNT int the form $$\pi\left(x\right)=\frac{x}{\log\left(x\right)}+O\left(\frac{x}{\log^{2}\left(x\right)}\right) $$ we have $$\sum_{p\leq N}\frac{1}{p}=\frac{\pi\left(N\right)}{N}+\int_{2}^{N}\frac{\pi\left(t\right)}{t^{2}}dt $$ $$=\frac{1}{\log\left(N\right)}+\int_{2}^{N}\frac{1}{t\log\left(t\right)}dt+O\left(\frac{1}{\log^{2}\left(N\right)}\right)+O\left(\int_{2}^{N}\frac{1}{t\log^{2}\left(t\right)}dt\right) $$ so...