Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In David M. Burton's book on Elementary Number Theory I have found the following words,

... The first demonstrable progress toward comparing $\pi(x)$ with $\dfrac {x}{\ln x}$ was made by ... P. L. Tchebycheff..., he proved that there exist positive constants $a$ and $b$ with $a$ $<$ $1$ $<$ $b$ such that- $$\dfrac{ax}{\ln x} < \pi(x) < \dfrac{bx}{\ln x}$$

For all sufficiently large $x$

Tchebycheff also showed that if there exists a limit of $\displaystyle\dfrac{\pi(x)}{\dfrac {x}{\ln x}}$ then it must be $1$.

Suppose now I have proved that the sequence $u_n$ $=$ $\displaystyle\dfrac{\pi(n)}{\dfrac {n}{\ln n}}$ is strictly decreasing for all sufficiently large $n$ and therefore has a limit (since a strictly decreasing sequence of positive numbers which is bounded below has a limit, more precisely its infimum). If I now attempt to conclude that since by Tchebycheff's result if there exists a limit of $\displaystyle\dfrac{\pi(x)}{\dfrac {x}{\ln x}}$ then it must be $1$, we must have, $$\lim_{n \to \infty} \dfrac{\pi(n)}{\dfrac {n}{\ln n}} = 1$$

Where would I be wrong?

Noting that the function $f(x)= \dfrac {x}{\ln x}$ is strictly increasing, I now intend to give a proof of the result that the sequence $u_n$ $=$ $\displaystyle\dfrac{\pi(n)}{\dfrac {n}{\ln n}}$ is strictly decreasing for all sufficiently large $n$.

We proceed by considering the following cases,

Case 1

In this case our assumption will be $\pi(n+1)=\pi(n)$.

It immediately follows that $u_n$ $>$ $u_{n+1}$.

Case 2

Now assume that $\pi(n+1)=\pi(n)+1$.

Notice that,

$$\dfrac{\pi(n)}{\dfrac{n}{\ln n}} > \dfrac{\pi(n)+1}{\dfrac{n+1}{\ln (n+1)}} \implies \dfrac{\pi(n)(n+1)}{\ln (n+1)} > \dfrac{n(\pi(n)+1)}{\ln n}$$ Which in turn implies that, $$n^{\displaystyle(1+\dfrac{1}{n})(1-\dfrac{1}{\pi(n)+1})}>(n+1)$$ which is obvious for all sufficiently large $n$.

share|cite|improve this question
Sounds reasonable. But I do not believe the sequence is ultimately strictly decreasing. – André Nicolas Jun 16 '14 at 6:00
The sequence is ultimately strictly decreasing. It can be proved by using the inequality $\pi(x)>\ln \ln x$ whose proof, of course doesn't depend on the proof of Prime Number Theorem. This is also the reason for the bound on $n$. – William Hilbert Jun 16 '14 at 6:04
@WilliamHilbert: How do you prove it is strictly decreasing using only that inequality? Obviously f(3) could be $10^{80}$, f(4) could be 3, and f(5) could be $10^{80}$ again and f(x) could obey that inequality. So what other behavior are you assuming? – ex0du5 Jun 16 '14 at 6:26
The inequality $$\dfrac{\pi(n)}{\dfrac{n}{\ln n}} > \dfrac{\pi(n)+1}{\dfrac{n+1}{\ln (n+1)}}$$ doesn't hold, as had been told in Case 2 of the proof. Silly mistake! – William Hilbert Jun 16 '14 at 12:30

The sequence $u_n = \frac{\pi(n)}{ \frac{n}{\ln n} }$ is not strictly decreasing for all $n > e^{e^2}$. This is straightforward to check using WolframAlpha. We have $1619 > e^{e^2}$ and $u_{1619} \approx 1.168$ while $u_{1621} \approx 1.171$. More generally I would guess that for $p$ a sufficiently large prime $u_p$ is greater than $u_{p-1}$, so the sequence should never be eventually strictly decreasing.

share|cite|improve this answer
I will be glad if you would produce a proof of your claim because, I also think that a proof Prime Number Theorem will not be so easy and there must be some mistake in my argument. – William Hilbert Jun 16 '14 at 12:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.