4
$\begingroup$

In number theory the three Mertens' theorems are the following.

Mertens' $1$st theorem. For all $n\geq2$

$$\left\lvert\sum_{p\leqslant n} \frac{\ln p}{p} - \ln n\right\rvert \leq 2.$$

Mertens' $2$nd theroem.

$$\lim_{n\to\infty}\left(\sum_{p\le n}\frac1p -\ln\ln n-M\right) =0,$$

where $M$ is the Meissel–Mertens constant.

Mertens' $3$rd theroem.

$$\lim_{n\to\infty}\ln n\prod_{p\le n}\left(1-\frac1p\right)=e^{-\gamma},$$

where $\gamma$ is the Euler–Mascheroni constant.

What connections are there between in this three theorems, beside that all of them about prime series and products? What relationships are behind the scenes? I know that the $2$nd theorem is connected to prime number theorem, and the other two theorems?

The motivation of the question is this really nice answer.

$\endgroup$
4
$\begingroup$

Mertens' third theorem is just the exponentiated version of the second theorem (without the bounds that Mertens proved for his second theorem):

\begin{align} -\ln\Biggl(\ln n\prod_{p\leqslant n}\biggl(1 - \frac{1}{p}\biggr)\Biggr) &= -\ln \ln n - \sum_{p\leqslant n} \ln \biggl(1 - \frac{1}{p}\biggr)\\ &= \Biggl(\sum_{p\leqslant n}\frac{1}{p} - \ln \ln n - M\Biggr) + \Biggl(M - \sum_{p\leqslant n} \biggl(\ln\biggl(1-\frac{1}{p}\biggr) + \frac{1}{p}\biggr)\Biggr), \end{align}

where the first term converges to $0$ by Mertens' second theorem, and the second term converges to $\gamma$ by definition of $M$.

Mertens' bounds in the second theorem and estimates for

$$\sum_{p > n}\biggl(\ln\biggl(1-\frac{1}{p}\biggr)+\frac{1}{p}\biggr)$$

give you bounds for

$$e^\gamma\ln n\prod_{p\leqslant n}\biggl(1-\frac{1}{p}\biggr),\tag{$\ast$}$$

and conversely bounds for that give you bounds for

$$\left\lvert\sum_{p\leqslant n}\frac{1}{p} - \ln \ln n - M\right\rvert,\tag{$\ast\!\ast$}$$

but it is doubtful whether one can directly prove bounds for $(\ast)$ that give you back Mertens' bounds for $(\ast\ast)$.

One can use Mertens' first theorem to derive the second via an integration by parts, Hardy and Wright for example do that, but don't give explicit bounds on $(\ast\ast)$.

For $x > 0$ we define

$$S(x) := \sum_{p\leqslant x} \frac{\ln p}{p}.$$

Mertens' first theorem tells us

$$\lvert S(x) - \ln x\rvert \leqslant 2 + O(x^{-1}),$$

and we can write

$$T(x) := \sum_{p\leqslant x} \frac{1}{p} = \int_{3/2}^x \frac{1}{\ln t}\,dS(t)$$

with a (Riemann/Lebesgue-) Stieltjes integral. Integration by parts yields

\begin{align} T(x) &= \int_{3/2}^x \frac{1}{\ln t}\,dS(t)\\ &= \frac{S(x)}{\ln x} - \frac{S(3/2)}{\ln \frac{3}{2}} - \int_{3/2}^x S(t)\,d\biggl(\frac{1}{\ln t}\biggr)\\ &= \frac{S(x)}{\ln x} + \int_{3/2}^x \frac{S(t)}{t(\ln t)^2}\,dt\\ &= \frac{S(x)}{\ln x} + \int_{3/2}^x \frac{dt}{t\ln t} + \int_{3/2}^x \frac{S(t) - \ln t}{t(\ln t)^2}\,dt\\ &= \ln \ln x + \underbrace{1 - \ln \ln \frac{3}{2} + \int_{3/2}^\infty \frac{S(t) - \ln t}{t(\ln t)^2}\,dt}_M + \underbrace{\frac{S(x)-\ln x}{\ln x} - \int_x^\infty \frac{S(t)-\ln t}{t(\ln t)^2}\,dt}_{O\bigl(\frac{1}{\ln x}\bigr)}. \end{align}

I'm not sure, however, whether one can get exactly Mertens' bounds on $(\ast\ast)$ easily from that.

So in a way, Mertens' first theorem is the most powerful, since it implies the others, at least if we don't need explicit bounds for the differences.

$\endgroup$
  • $\begingroup$ Thank you for your answer. +1. Hardy and Wright means this: Hardy, G. H. and Wright, E. M. An Introduction to the Theory of Numbers, 5th ed. Oxford, England: Oxford University Press, p. 351, 1979. ? Furtnermore, Mertens gave explicit bounds on $(**)$, not? This bound: $$\frac 4{\ln(n+1)} +\frac 2{n\ln n}$$ $\endgroup$ – user153012 Nov 26 '14 at 20:13
  • 1
    $\begingroup$ Yes, exactly, pp. 350/351, even the edition is the same. $\endgroup$ – Daniel Fischer Nov 26 '14 at 20:15
  • 1
    $\begingroup$ Yes, Mertens gave explicit bounds, Hardy/Wright didn't. If you take care, I expect you can get at least similar bounds from the integration by parts method, but I have no idea how hard it would be. Mertens' original proof proceeds in a different way, it is longer and more involved. I have to confess that before I could understand it, I would have to write it on paper and translate it into terminology I can better understand than Mertens' terminology. $\endgroup$ – Daniel Fischer Nov 26 '14 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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