Psi and theta differ by at most a square-root term

Subtracting the $k=1$ term from [L1] gives $\psi (x)-\theta (x)={\sum }_{k\ge 2}\theta (x^{1\mathrm{/}k})\mathrm{.}$ Every summand is nonnegative, so $0\le \psi (x)-\theta (x)\mathrm{.}$

The $k=2$ term is $\theta (\sqrt{x})\le \sqrt{x}\log x$, because there are at most $\sqrt{x}$ primes at most $\sqrt{x}$, and each contributes at most $\log \sqrt{x}\le \log x$. The terms with $k>\lfloor {\log }_{2}x\rfloor$ vanish because $x^{1\mathrm{/}k}<2$. For $3\le k\le \lfloor {\log }_{2}x\rfloor$, one has $x^{1\mathrm{/}k}\le x^{1\mathrm{/}3}$, so ${\sum }_{k=3}^{\lfloor {\log }_{2}x\rfloor }\theta (x^{1\mathrm{/}k})\le x^{1\mathrm{/}3}\log x\cdot \lfloor {\log }_{2}x\rfloor \le K_1^{\mathrm{\prime }}\sqrt{x}\log x$ for a fixed constant $K_1^{\mathrm{\prime }}$, because $(\log x)x^{-1\mathrm{/}6}$ is bounded for $x\ge 2$. Together with step 1.1, this proves $\psi (x)-\theta (x)\le K_1\sqrt{x}\log x$ for a suitable constant $K_1$.

By [L2], choose $C>0$ and $y_0\ge 2$ such that $\theta (y)\le Cy$ for every $y\ge y_0$. Put $C^{\mathrm{\prime }}:=\max \{C,\theta (y_0)\}$. If $1\le y\le y_0$, monotonicity gives $\theta (y)\le \theta (y_0)\le C^{\mathrm{\prime }}\le C^{\mathrm{\prime }}y$, while for $y\ge y_0$ one has $\theta (y)\le Cy\le C^{\mathrm{\prime }}y$. Thus $\theta (y)\le C^{\mathrm{\prime }}y$ for every real $y\ge 1$. Then for all sufficiently large $x$, the term $k=2$ satisfies $\theta (\sqrt{x})\le C^{\mathrm{\prime }}\sqrt{x}\mathrm{.}$ Also, using the finite range from step 2.1, ${\sum }_{k=3}^{\lfloor {\log }_{2}x\rfloor }\theta (x^{1\mathrm{/}k})\le C^{\mathrm{\prime }}{x}^{1\mathrm{/}3}\lfloor {\log }_{2}x\rfloor \le C^{\mathrm{\prime }}\sqrt{x}$ for large $x$, because $(\log x)x^{-1\mathrm{/}6}\to 0$. Therefore $\psi (x)-\theta (x)\le K_2\sqrt{x}$ for a suitable constant $K_2$.