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Let $\psi(x) := \sum_{n\leq x} \Lambda(n)$ where $\Lambda(n)$ is the Von-Mangoldt function. I want to show that if $$ \lim_{x \rightarrow \infty} \frac{\psi(x)}{x} =1 $$ then also $$\lim_{x\rightarrow \infty} \frac{\pi(x) \log x }{x}=1.$$

I tried to play a little bit with $\psi$, what I want to show is that:

$$\left| \frac{\pi(x) \log x}{x} -1 \right| \leq \left| \frac{\psi(x)}{x} -1 \right| \rightarrow 0$$

So I tried to develop $\psi$ a little bit, but I got astray.

So I have $$ \frac{\psi(x)}{x} -1 = \sum_{p^k \leq x , k \geq 1} \frac{\log p}{x} -1 = \frac{1}{x}\left(\sum_{p\leq x} \log p + \sum_{p^2\leq x} \log p + ...+ \sum_{p^k \leq x, p^{k+1} >x} \log p \right) -1 $$ and I want to estimate its aboslute value from below, but I don't have any idea?

Any hints?

Thanks.

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First instead of $\psi(x)$, we will deal in terms of $\theta(x)$ because of the following reason.

Clearly $$\psi(x) \geq \theta(x).$$ Further, $$ \begin{align*} \psi(x) & = \sum_{n \leq x} \Lambda(n) = \sum_{\overset{p^k \leq x}{p \text{ is a prime and }k \in \mathbb{Z}^+}} \log(p)\\ & = \sum_{\overset{p \leq x}{p \text{ is a prime}}} \log(p) + \sum_{\overset{p^2 \leq x}{p \text{ is a prime}}} \log(p) + \sum_{\overset{p^3 \leq x}{p \text{ is a prime}}} \log(p) + \cdots\\ & = \theta(x) + \theta(x^{1/2}) + \theta(x^{1/3}) + \theta(x^{1/4}) \cdots\\ & \leq \theta(x) + \frac{x^{1/2} \log(x)}{2} + \frac{x^{1/3} \log(x)}{3} + \frac{x^{1/4} \log(x)}{4} + \cdots\\ & = \theta(x) + \log(x) \left( \sum_{k=2}^{\lfloor \log_2(x) \rfloor} \frac{x^{1/k}}{k} \right)\\ & \leq \theta(x) + x^{1/2} \log(x) \left( \sum_{k=2}^{\lfloor \log_2(x) \rfloor} \frac1k \right)\\ & = \theta(x) + \mathcal{O} \left(x^{1/2} \log(x) \log(\log(x))\right) \end{align*} $$

Hence, if $\displaystyle \lim_{x \rightarrow \infty} \frac{\psi(x)}{x} = 1$, then so is $\displaystyle \lim_{x \rightarrow \infty} \frac{\theta(x)}{x} = 1$.

Further $\displaystyle \theta(x) = \sum_{\overset{p \leq x}{p \text{ is a prime}}} \log(p) = \sum_{2 \leq n \leq x} \log(n) (\pi(n) - \pi(n-1))$ i.e.

$$d \theta (x) = \log(x) d\pi(x)$$ Hence, we have that $\displaystyle d \pi(x) = \frac{d \theta(x)}{\log(x)}$. Since $\theta(x) = x + o(x)$ we get that $$\pi(x) = \int_{2^-}^{x} \frac{d \theta(t)}{\log(t)} = \int_{2^-}^{x} \frac{d t}{\log(t)} + \mathcal{O} \left( \int_{2^-}^{x} \frac{dt}{\log^3(t)} \right)= \frac{x}{\log(x)} + \mathcal{O} \left( \frac{x}{\log^3(x)} \right).$$

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  • $\begingroup$ Can somebody elaborate how the term $\mathcal{O} \left( \int_{2^-}^{x} \frac{dt}{\log^3(t)} \right)$ is obtained in the final line? $\endgroup$ – wiskundeliefhebber Jun 3 '15 at 13:55
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This is a modification of a proof given in Chandrasekharan's Introduction to Analytic Number Theory (p.65, Theorem 2). Since you ask for hints I'll try to leave out some of the details and retain the landmarks.

As you noted, we have

$$ \psi(x) = \sum_{p\leq x} \log p + \sum_{p^2\leq x} \log p + \sum_{p^3\leq x} \log p + \cdots, $$

the chain of sums being finite for $x > 0$ since $p^m \geq 2^m > x$ when $m > \log x/\log 2$. Now, if $x \geq 1$ and $p^n \leq x < p^{n+1}$, then $\log p$ occurs exactly $n$ times in the sum, and $n = \lfloor \log x / \log p \rfloor$. Thus

$$ \psi(x) = \sum_{p \leq x} \left\lfloor\frac{\log x}{\log p}\right\rfloor\log p. $$

The result then follows in two steps. First, show that

$$ \psi(x) \leq \pi(x) \log x. $$

Next, let $0 < \alpha < 1$, and show that

$$ \psi(x) \geq \sum_{x^\alpha < p \leq x} \left\lfloor\frac{\log x}{\log p}\right\rfloor\log p \geq \alpha \log x (\pi(x) - x^\alpha), $$

then divide by $x$ and let $x \to \infty$.

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