# Asymptotic behavior of Chebyshev functions

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?

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