Some Steps in the Proof of Theorem 4.1

$$\begin{array}{rl}\psi (x)& =\sum _{m\le {\log }_{2}x}\vartheta (x^{1/m})\\ & =\vartheta (x)+\sum _{2\le m\le {\log }_{2}x}\vartheta (x^{1/m}).\end{array}$$

Subtracting $\vartheta (x)$ from both sides, we get

$$\begin{array}{rl}\psi (x)-\vartheta (x)& =\sum _{2\le m\le {\log }_{2}x}\vartheta (x^{1/m})\\ & \le \sum _{2\le m\le {\log }_{2}x}{x}^{1/m}\log x^{1/m}\\ & \le \sum _{2\le m\le {\log }_{2}x}{x}^{1/2}\log x^{1/2}\\ & \le ({\log }_{2}x-1)x^{1/2}\log x^{1/2}\\ & \le ({\log }_{2}x)\sqrt{x}\log \sqrt{x}.\end{array}$$