Asking for help in proving a result related to Mangoldt function in number theory

Question

I am trying exercises of ch-4 of Apostol Introduction to analytic number theory and in 1 question I am struck. This question uses following condition

Can it be proved that $\sum_{n\leq x}\frac {\Lambda(n) } { n} $ = log (x) +O(1) ?

Due to Mangoldt function sum will be calculated only on prime values but I am not able to get desired result. Can you please help?

Answer 1

$$\log(n!) = n\log n+O(n)$$ $$=\sum_{p^k \le n} \lfloor n/p^k \rfloor\log p=n \sum_{p^k \le n} \frac{\log p}{p^k}+O(\sum_{p^k \le n} \log p) $$ Where $\sum_{p^k \le n} \log p=O(n)$ from $\log 4^n \sim \log {2n\choose n}\ge \log \prod_{p\in (n,2n]} p$.

Answer 2

Using the fact that $$n!=\prod_{p\;\mathrm{prime}} p^{j(n,p)}=\prod_{\substack{p\;\mathrm{prime}\\p\leq n}} p^{j(n,p)}\quad\mathrm{where}\quad j(n,p)=\sum_{m=1}^{\infty} \bigg\lfloor \frac{n}{p^m}\bigg\rfloor$$ Therefore we obtain $$\sum_{n=2}^{\lfloor x\rfloor} \ln(n)=\ln\Bigg(\prod_{\substack{p\;\mathrm{prime}\\p\leq x}} p^{j(\lfloor x\rfloor,p)}\Bigg)=\sum_{\substack{p\;\mathrm{prime}\\p\leq x}} j(\lfloor x\rfloor,p)\ln(p)=\sum_{\substack{p\;\mathrm{prime}\\p^m\leq x}} \bigg\lfloor \frac{x}{p^m}\bigg\rfloor\ln(p)=\sum_{n\leq x} \bigg\lfloor \frac{x}{n}\bigg\rfloor\Lambda(n)$$ since $\ln(1)=0$ and for any real number $x$ and any positive integer $n$, we have $\lfloor\lfloor x\rfloor/n\rfloor=\lfloor x/n\rfloor$. Now we use the equivalent statement of the Prime Number Theorem, namely the asymptotic formula for Chebyshev's $\psi$-function $$\psi(x)=\sum_{n\leq x} \Lambda(n)=O(x)$$ If we remove the floor function in the last sum, we have \begin{align*}\begin{split} &\sum_{n\leq x} \frac{x}{n}\Lambda(n)<\sum_{n\leq x} \bigg\lfloor \frac{x}{n}\bigg\rfloor\Lambda(n)+\sum_{n\leq x}\Lambda(n)=\sum_{n\leq x} \ln(n)+O(x)\\ &\Leftrightarrow \sum_{n\leq x} \frac{x}{n}\Lambda(n)=\sum_{n\leq x} \ln(n)+O(x) \end{split}\end{align*} since for any real number $x$, we have $x<\lfloor x\rfloor+1$. Now we use the asymptotic formula for $\sum_{n\leq x} \ln(n)=x\ln(x)+O(x)$, therefore we obtain $$\sum_{n\leq x} \frac{x}{n}\Lambda(n)=x\ln(x)+O(x)$$ Dividing by $x$ on both sides and it follows.