Lower bound for the limit of a ratio related to the distribution of primes.

Question

Let $\theta(x)=\sum_{p\leq x}\ln{p}$ denote the first Chebyshev function and let $f(x):=\sum_{p\leq x}p$. I'm hoping to show $$\liminf_{x\to\infty}\frac{\theta(x)^2}{f(x)\ln{f(x)}}\geq1.$$ That is, that for every $x_0>0$ and $\varepsilon>0$ there exists $x>x_0$ such that $$\frac{\theta(x)^2}{f(x)\ln{f(x)}}>1-\varepsilon.$$ I've tried to find bounds on $\theta(x)$ and $f(x)$ using the prime number theorem, for example $$\theta(x)\geq c(x)x,\qquad\text{ and }\qquad f(x)\leq d(x)\frac{x^2}{\ln{x}},$$ where $c(x)$ and $d(x)$ tend to $1$ as $x$ tends to infinity. There are many such $c(x)$ and $d(x)$, but to make my argument work I need that $$\lim_{x\to\infty}\frac{c(x)^2}{d(x)}\geq2,$$ but I am unable to find such $c(x)$ and $d(x)$. Do they even exist, or is what I want to show actually false?

Answer 1

The prime number theorem asserts $\theta(x) \sim x$, so for every $\epsilon > 0$, $\theta(x) \ge (1-\epsilon)x$ for $x$ large enough. This gives $\theta(x)^2 \ge (1-\epsilon)^2x^2$. If we can show $f(x) \le (\frac{1}{2}+\epsilon)\frac{x^2}{\ln(x)}$ for large $x$, then $\ln f(x) \le 2\ln(x)+\ln(\frac{1}{2}+\epsilon)-\ln\ln(x)$ and so $f(x)\ln f(x) \le (1+3\epsilon)x^2$ for $x$ large enough, since the $x^2$ term dominates. So, for large enough $x$, $\frac{\theta(x)}{f(x)\ln f(x)} \ge \frac{(1-\epsilon)^2}{1+3\epsilon}$ which will give you what you want.

By summation by parts, $f(x) = \pi(x)x-\int_2^x \frac{t}{\ln t}dt \le \frac{(1+\epsilon)x^2}{\ln(x)}-\int_2^x \frac{t}{\ln t}dt$, and $\int_2^x \frac{t}{\ln t}dt \ge \int_{\sqrt{x}}^x \frac{t}{\ln(t)} \ge \frac{1}{\ln(x)}[\frac{1}{2}x^2-\frac{1}{2}\sqrt{x}^2] \ge (\frac{1}{2}-\epsilon)\frac{x^2}{\ln(x)}$, so $f(x) \le (\frac{1}{2}+2\epsilon)\frac{x^2}{\ln(x)}$, as desired.