Let $\vartheta(x)=\sum_{p\le x} \log p$, $ \psi(x) = \sum_{p^k\le x}\log p$ be chebyshev functions. I want to show that if $\vartheta(x) \sim x$, then $\psi(x) \sim x$. ($f(x)\sim g(x)$ means that $\lim_{x\to \infty}f(x)/g(x)=1$)

I know that $\psi(x)=\sum_{n=1}^\infty \vartheta \left(x^{1/n}\right)$, so $$\lim_{x \to \infty} \frac{\psi(x)}{x}=\lim_{x \to \infty}\sum_{n=1}^\infty \frac{\vartheta \left(x^{1/n}\right)}{x}$$

But I'm not sure the interchange of limit and sigma is possible. Is it possible? Or is there other methods to solve this problem?


One method (the one found in Apostol's Introduction to Analytic Number Theory) is to first show that

$$ \psi(x) = \sum_{n \leq \log_2 x} \vartheta\left(x^{1/n}\right), $$

then show that

$$ \vartheta(y) \leq y\log y $$

and apply this to the above identity.


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