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.
Best Answer
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.