I'm trying to prove that $$\sum_{n\le x}(\psi(\frac xn)-\vartheta(\frac xn))\Lambda (n)=O (x) $$
After some calculations, l arrived to $$O (\sqrt{x}\log x\sum_{m=1}^\infty x^{\frac 1m} (\frac{1}{\sqrt{2}})^m)$$
So l need to evaluate that sum. Could anyone help me, please?
Remark: This question is related to problem 4.22 in Apostol's book. The problem is requested to prove that Selberg's asymptotic formula is equivalent to
$$\vartheta (x)\log x+\sum_{p\le x}\vartheta (\frac xp)\log p=2x\log x+O (x) $$
My idea was to first show that
$$\vartheta (x)\log x+\sum_{n\le x}\vartheta (\frac xn)\Lambda (n)=2x\log x+O (x) $$
Thus, i've needed to show that
$$\sum_{n\le x}(\psi(\frac xn)-\vartheta(\frac xn))\Lambda (n)=O (x) $$
$$\sum_{n\le x}(\psi(\frac xn)-\vartheta(\frac xn))\Lambda (n)=O(\sum_{n \le x}(\frac xn)^{1/2} (\psi(n+1)-\psi(n)))\\ = O(\psi(x)+\sum_{n \le x}((\frac xn)^{1/2}-(\frac x{n+1})^{1/2}) \psi(n)) $$
$$ = O(x+\sum_{n \le x}((\frac xn)^{1/2}-(\frac x{n+1})^{1/2}) n)= O(x+\sum_{n \le x} (\frac xn)^{1/2}) = O(x)$$
where I used $\psi(x)-\psi(x/2) = O(\log {x\choose x/2} ) = O(x)$,
$\psi(x) = O(x)$, $\psi(x)-\vartheta(x) = O(x^{1/2})$ and partial summation.