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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 $n$.

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  • $\begingroup$ Sounds reasonable. But I do not believe the sequence is ultimately strictly decreasing. $\endgroup$ Jun 16, 2014 at 6:00
  • $\begingroup$ 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 $n$. $\endgroup$ Jun 16, 2014 at 6:04
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    $\begingroup$ @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? $\endgroup$
    – ex0du5
    Jun 16, 2014 at 6:26
  • $\begingroup$ 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! $\endgroup$ Jun 16, 2014 at 12:30

1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$ Jun 16, 2014 at 12:17

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