I have a problem...
If $A(x)=\sum_{n\leq x}a(n)$ where $a(n)=\frac{1}{k}$ if $n=p^{k}$ and $a(n)=0$ in other case, show that $A(x)=\pi(x)+O(\sqrt{x}\:log\:logx)$
I think I should use the theorem of prime numbers, but I don't know how. \\ is it possible to do it without the theorem?
$$A(x)=\sum_{p^k\leq x}\frac{1}{k}=\sum_{p\leq x}1+\sum_{p^2\leq x}\frac{1}{2}+\sum_{p^3\leq x}\frac{1}{3}\cdots$$ $$=\sum_{p\leq x}1+\sum_{p\leq x^{1/2}}\frac{1}{2}+\sum_{p\leq x^{1/3}}\frac{1}{3}+\cdots$$ $$=\pi(x)+\frac{1}{2}\pi(x^{1/2})+\frac{1}{3}\pi(x^{1/3})+\cdots$$ $$=\pi(x)+\frac{1}{2}\pi(x^{1/2})+\sum_{3\leq k \leq \log_2(x)}\frac{1}{k}\pi(x^{1/k})$$ $$=\pi(x)+\frac{1}{2}\pi(x^{1/2})+O(x^{1/3}\ln(x))$$ $$=\pi(x)+O(x^{1/2})=\pi(x)+O(x^{1/2}\ln(\ln(x)))$$