# Show that $\limsup_{x \to \infty} \frac{\pi(x)}{x/ \log x} \geqslant 1.$

Show that $$\displaystyle\limsup_{x \to \infty} \dfrac{\pi(x)}{x/ \log x} \geqslant 1.$$

I've seen $$\displaystyle\lim_{x \to \infty}$$ operator, but I haven't seen $$\displaystyle\limsup_{x \to \infty}$$ or $$\displaystyle\liminf_{x \to \infty}$$ before. What do they mean exactly? Based on the image from wikipedia:

It seems that we need $$\displaystyle\limsup_{x \to \infty}$$ and $$\displaystyle\liminf_{x \to \infty}$$ when a sequence is not strictly increasing or decreasing, but jumping between values. I've not taken analysis, so these terms are not so familiar to me.

$$\textbf{Edit:}$$ These concepts make sense to me now.

The following was proved in my notes:

For every real number $$y \geqslant 2,$$

$$\displaystyle\sum\limits_{p \leqslant y} \dfrac{1}{p} > \log(\log y) - 1.$$

This is supposed to be useful in application in this problem, but I am not sure how exactly.

The prime number theorem states that $$\displaystyle\lim_{x \to \infty} \dfrac{\pi(x)}{x/ \log x} = 1,$$ but we didn't prove it in my textbook or notes, so I'd like to refrain from using it if necessary.

• $\sup$ and $\inf$ are used to denote supremum and infimum of a set respectively. Commented Apr 16, 2015 at 7:13
• Summation by parts. Assume that $\limsup\limits \frac{\pi(x)}{x/\log x} < 1$. Then there is a $c < 1$ and an $x_0$ such that for $x \geqslant x_0$ you have $\pi(x) \leqslant c \frac{x}{\log x}$. A summation by parts shows that for all large enough $y$ it follows that $$\sum_{p \leqslant y} \frac{1}{p} < \frac{1 + c}{2} \log \log y,$$ which contradicts the lower bound from your notes. Commented Mar 3, 2018 at 23:05

This is a classical result of a course in analytic number theory. Chebyshev first showed that $$c_1\frac{x}{\log(x)}\le \pi(x)\le c_2\frac{x}{\log(x)}$$ for all $x\ge 96098$, with the Chebyshev constants \begin{align*} c_1 & = \log(2^{1/2}3^{1/3}5^{1/5}30^{-1/30})\approx 0.921292022934, \\[0.3cm] c_2 & =\frac{6}{5}c_1\approx 1.10555042752. \\ \end{align*} He proved then the following result:
Proposition(Chebychev): We have $$4\log(2)\ge \limsup\limits_{x\rightarrow \infty}\frac{\pi(x)}{x/\log (x)}\ge \liminf\limits_{x\rightarrow \infty}\frac{\pi(x)}{x/\log (x)}\ge \log(2).$$ The upper and lower bounds were finally improved to $1$, implying PNT (Jacques Hadamard and Charles de la Vallée-Poussin). For a reference, e.g., see here.
Use $$\sum_{p \leq x} \frac{1}{p}= \frac{\pi(x)}{x}+ \int_2^x \frac{\pi(y)}{y^2} dy,$$(this can be verified easily,e.g. Show that $\sum\limits_{p \leqslant x}1/p = \frac{\pi(x)}{x} + \int_2^x \frac{\pi(u)}{u^2} du.$) and prove by contradiction. By some simple calculation, when $x$ is sufficiently large, RHS of the above equality becomes strictly smaller than $log log x-1$, contradicting to the given fact $$\sum_{p \leq x} \frac{1}{p} >log log x −1.$$