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Added (December, 2017). I came upon an observation giving also a 'trivial' proof of the reverse elementary implication of the two purely qualitative forms, multiplicative and logarithmic, of the prime number theorem: $\psi(X) \sim X \Leftrightarrow S(X) = \log{X} - \gamma + o(1)$. The following seems to have been missed in the literature on elementary methods which, at this point, seem all to quote a somewhat more involved Tauberian theorem of Axer; cf. section 8.1.1 of Montgomery and Vaughan's book (Multiplicative Number Theory: I) or, for a more general setting, chapter 14 of Diamond and Zhang's recent book on Beurling Generalized Numbers (really this paper of theirs). The simpler argument below also extends easily to number fields, supplying a particularly easy proof of the 'elementary equivalence' of Landau's prime ideal theorem and number field sharp Mertens. Incidentally, as I happen to recall, this addresses a slightly curious point that had come up in the comments to this answer of Eric Naslund. Remembering also my answer here, I figured it may be worth to record the following observation as an addendum to it, sticking for simplicity to the rational case assumed in this question.

A proof of $\psi(X) \sim X \Rightarrow S(X) = \log{X} - \gamma + o(1)$. For simplicity, let me stick to $\mathbb{Q}$. The case of a number field $K$ has the same result with $\gamma$ generalized as the 'Euler-Kronecker invariant' $\gamma_K$.

The key is to observe that the formula $$ X^{-1} \log{X!} = \sum_{n \leq X/T} \frac{\Lambda(n)}{n} + \sum_{m \leq T} \frac{1}{X} \Big( \psi\Big( \frac{X}{m} \Big) - \psi\Big( \frac{X}{T} \Big) \Big) + O(1/T) $$ holds uniformly in the two parameters $X, T \geq 1$, with an absolute implied coefficient. It interpolates between Mertens's estimate (case $T = 1$) and Chebyshev's convolution formula $\log{X!} = \sum_m \psi(X/m)$ (case $T = \infty$). But the general formula also follows, after a moment of reflection, from Chebyshev's argument with the prime factorization of $X!$. Divide the moduli into the ranges $n \leq X/T$ and $n > X/T$. The total contribution of the latter are exactly accounted for by the second sum. For a small modulus $n \leq X/T$, the contribution via the prime factorization is $X^{-1} \lfloor X/n \rfloor \Lambda(n) = \frac{\Lambda(n)}{n} + O\Big(\frac{\Lambda(n)}{X}\Big)$, neglecting the fractional part. The $O(1/T)$ term then comes from summing these for $n \leq X/T$, and using Chebyshev's estimate $\sum_{n \leq Y} \Lambda(n) \ll Y$. (In the number field generalization, the latter estimates extend as lattice point counts.)

Now, by Stirling's asymptotic, the qualitative $\psi(X) \sim X \Rightarrow S(X) = \log{X} - \gamma + o(1)$ implication is immediate from the observed formula upon first letting $X \to \infty$ and then $T \to \infty$.

Because the scale is too small in Mertens's theorem, and the prime number theorem as well as the Riemann hypothesis are hidden by the $O(1/\log{X})$ notation.

Indeed, the former amounts to strengthening this term to $o(1/\log{X})$; the latter - to $O(1/\sqrt{X})$.

One could go for equivalent statements to a still smaller scale. What does an estimate $\sum_{p < X} 1/p^2 = \mathrm{const} + O(1/X)$ tell you about the prime numbers? Absolutely nothing. Yet, here too, one could if one wished express the prime number theorem as an estimate on the $O(1/X)$ term. (Let me not elaborate on this. )

The prime number theorem is a statement about convergence, about an oscillatory term - in Mertens's case, $\log{X}$ times the $O(1/\log{X})$ remainder term - going to zero. The emphasis here is in the existence of a limit. Chebyshev knew that if $\psi(X)/X-1$ converges then it does to $0$; likewise, the exact value of the constant $M$ here is of no significance for the prime number theorem as soon as one can handle the oscillating term.

Let me try to clarify this by going to a slightly bigger scale where the prime number theorem begins to emerge outside of $o(1)$. This is also more natural - indeed it was how Mertens's theorem was proved.

By partial summation, Mertens's estimate is equivalent to $\sum_{p < X} (\log{p})/(p-1) = \log{X} + O(1)$; or, if one prefers, $\sum_{n < X} \Lambda(n)/n = \log{X} + O(1)$. The prime number theorem however is the statement that the $O(1)$ term converges to a constant: $\sum_{n < X} \Lambda(n)/n = \log{X} - \gamma + o(1)$. Indeed, the related bound $\sum_{n < X} \Big( \frac{1}{n} - \frac{1}{X} \Big) \Lambda(n) = \log{X} - 1 - \gamma + o(1)$, another form of the prime number theorem, is what de la Vallee Poussin actually obtained in his original paper. Here $\gamma = 0.57\ldots$ is Euler's constant, but this is of no importance for us, see the preceding and next paragraphs. Also the $\log{X}$ logarithmic pole corresponds to the pole of $\zeta(s)$ at $s=1$, whereas the $o(1)$ term expresses there being no zeros with $\mathrm{Re}(s) = 1$. The Riemann hypothesis is the correspondingly stronger bound $O(1/\sqrt{X})$ on the $o(1)$ oscillating term. At this intermediate scale, in contrast to $\psi(X) = X + O\big( \sqrt{X} (\log{X})^2 \big)$, a logarithmic factor in addition to the square root is not required, as $\sum_{\rho} 1 / |\rho|^2 < \infty$ over the zeros.

Here finally is how to deduce the more usual form $\psi(X) \sim X$ of the prime number theorem from the refinement $S(X) := \sum_{p < X} \Lambda(n)/n = \log{X} -\gamma + o(1)$ of Mertens's theorem: Summing by parts, $\psi(X) = XS(X) - \int_1^X S(t) \, dt = X(\log{X} - \gamma + o(1)) - \big( \int_1^X \log{t} \, dt - \gamma X + o(X) \big) = X + o(X)$.

Because the scale is too small in Mertens's theorem, and the prime number theorem as well as the Riemann hypothesis are hidden by the $O(1/\log{X})$ notation.

Indeed, the former amounts to strengthening this term to $o(1/\log{X})$; the latter - to $O(1/\sqrt{X})$.

[Incidentally, one could go for equivalent statements to a still smaller scale. Of course, an estimate like $\sum_{p < X} 1/p^2 = \mathrm{const} + O(1/X)$ does not say anything at all about the primes. But one could, if one wished, express the prime number theorem by elaborating on the $O(1/X)$ term. ]

To elaborate on this a bit, let me go to a slightly bigger scale where the prime number theorem begins to emerge outside of $o(1)$. This is also more natural; indeed, it was how Mertens's theorem was proved.

By partial summation, Mertens's estimate is equivalent to $\sum_{p < X} (\log{p})/(p-1) = \log{X} + O(1)$; or, if one prefers, $\sum_{n < X} \Lambda(n)/n = \log{X} + O(1)$. The prime number theorem however is the statement that the $O(1)$ term converges to a constant: $\sum_{n < X} \Lambda(n)/n = \log{X} - \gamma + o(1)$. Indeed, the related bound $\sum_{n < X} \Big( \frac{1}{n} - \frac{1}{X} \Big) \Lambda(n) = \log{X} - 1 - \gamma + o(1)$, another form of the prime number theorem, is what de la Vallee Poussin actually obtained in his original paper. Here $\gamma = 0.57\ldots$ is Euler's constant, but this is of no importance for us, see the next paragraph. Also the $\log{X}$ logarithmic pole corresponds to the pole of $\zeta(s)$ at $s=1$, whereas the $o(1)$ term expresses there being no zeros with $\mathrm{Re}(s) = 1$. The Riemann hypothesis is the correspondingly stronger bound $O(1/\sqrt{X})$ on the $o(1)$ oscillating term. At this scale, in contrast to $\psi(X) = X + O\big( \sqrt{X} (\log{X})^2 \big)$ or $\pi(X) = \mathrm{Li}(X) + O(\sqrt{X}\log{X})$, a logarithmic factor in addition to the square root is not required, as $\sum_{\rho} 1 / |\rho|^2 < \infty$ over the zeros.

Here finally is how to deduce the more usual form $\psi(X) \sim X$ of the prime number theorem from the refinement $S(X) := \sum_{p < X} \Lambda(n)/n = \log{X} -\gamma + o(1)$ of Mertens's theorem: Summing by parts, $\psi(X) = XS(X) - \int_1^X S(t) \, dt = X(\log{X} - \gamma + o(1)) - \big( \int_1^X \log{t} \, dt - \gamma X + o(X) \big) = X + o(X)$.

Because the scale is too small in Mertens's theorem, and the prime number theorem as well as the Riemann hypothesis are hidden by the $O(1/\log{X})$ notation.

Indeed, the former amounts to strengthening this term to $o(1/\log{X})$; the latter - to $O(1/\sqrt{X})$.

One could go for equivalent statements to a still smaller scale. What does an estimate $\sum_{p < X} 1/p^2 = \mathrm{const} + O(1/X)$ tell you about the prime numbers? Absolutely nothing. Yet, here too, one could if one wished express the prime number theorem as an estimate on the $O(1/X)$ term. (Let me not elaborate on this. )

The prime number theorem is a statement about convergence, about an oscillatory term - in Mertens's case, $\log{X}$ times the $O(1/\log{X})$ remainder term - going to zero. The emphasis here is in the existence of a limit. Chebyshev knew that if $\psi(X)/X-1$ converges then it does to $0$; likewise, the exact value of the constant $M$ here is of no significance for the prime number theorem as soon as one can handle the oscillating term.

Let me try to clarify this by going to a slightly bigger scale where the prime number theorem begins to emerge outside of $o(1)$. This is also more natural - indeed it was how Mertens's theorem was proved.

By partial summation, Mertens's estimate is equivalent to $\sum_{p < X} (\log{p})/(p-1) = \log{X} + O(1)$; or, if one prefers, $\sum_{n < X} \Lambda(n)/n = \log{X} + O(1)$. The prime number theorem however is the statement that the $O(1)$ term converges to a constant: $\sum_{n < X} \Lambda(n)/n = \log{X} - \gamma + o(1)$. Indeed, the related bound $\sum_{n < X} \Big( \frac{1}{n} - \frac{1}{X} \Big) \Lambda(n) = \log{X} - 1 - \gamma + o(1)$, another form of the prime number theorem, is what de la Vallee Poussin actually obtained in his original paper. Here $\gamma = 0.57\ldots$ is Euler's constant, but this is of no importance for us, see the preceding and next paragraphs. Also the $\log{X}$ logarithmic pole corresponds to the pole of $\zeta(s)$ at $s=1$, whereas the $o(1)$ term expresses there being no zeros with $\mathrm{Re}(s) = 1$. The Riemann hypothesis is the correspondingly stronger bound $O(1/\sqrt{X})$ on the $o(1)$ oscillating term. At this intermediate scale, in contrast to $\psi(X) = X + O\big( X (\log{X})^2 \big)$, a logarithmic factor is not required as $\sum_{\rho} 1 / |\rho|^2 < \infty$ over the zeros.

Here finally is how to deduce the more usual form $\psi(X) \sim X$ of the prime number theorem from the refinement $S(X) := \sum_{p < X} \Lambda(n)/n = \log{X} -\gamma + o(1)$ of Mertens's theorem: Summing by parts, $\psi(X) = XS(X) - \int_1^X S(t) \, dt = X(\log{X} - \gamma + o(1)) - \big( \int_1^X \log{t} \, dt - \gamma X + o(X) \big) = X + o(X)$.

Because the scale is too small in Mertens's theorem, and the prime number theorem as well as the Riemann hypothesis are hidden by the $O(1/\log{X})$ notation.

Indeed, the former amounts to strengthening this term to $o(1/\log{X})$; the latter - to $O(1/\sqrt{X})$.

One could go for equivalent statements to a still smaller scale. What does an estimate $\sum_{p < X} 1/p^2 = \mathrm{const} + O(1/X)$ tell you about the prime numbers? Absolutely nothing. Yet, here too, one could if one wished express the prime number theorem as an estimate on the $O(1/X)$ term. (Let me not elaborate on this. )

The prime number theorem is a statement about convergence, about an oscillatory term - in Mertens's case, $\log{X}$ times the $O(1/\log{X})$ remainder term - going to zero. The emphasis here is in the existence of a limit. Chebyshev knew that if $\psi(X)/X-1$ converges then it does to $0$; likewise, the exact value of the constant $M$ here is of no significance for the prime number theorem as soon as one can handle the oscillating term.

Let me try to clarify this by going to a slightly bigger scale where the prime number theorem begins to emerge outside of $o(1)$. This is also more natural - indeed it was how Mertens's theorem was proved.

By partial summation, Mertens's estimate is equivalent to $\sum_{p < X} (\log{p})/(p-1) = \log{X} + O(1)$; or, if one prefers, $\sum_{n < X} \Lambda(n)/n = \log{X} + O(1)$. The prime number theorem however is the statement that the $O(1)$ term converges to a constant: $\sum_{n < X} \Lambda(n)/n = \log{X} - \gamma + o(1)$. Indeed, the related bound $\sum_{n < X} \Big( \frac{1}{n} - \frac{1}{X} \Big) \Lambda(n) = \log{X} - 1 - \gamma + o(1)$, another form of the prime number theorem, is what de la Vallee Poussin actually obtained in his original paper. Here $\gamma = 0.57\ldots$ is Euler's constant, but this is of no importance for us, see the preceding and next paragraphs. Also the $\log{X}$ logarithmic pole corresponds to the pole of $\zeta(s)$ at $s=1$, whereas the $o(1)$ term expresses there being no zeros with $\mathrm{Re}(s) = 1$. The Riemann hypothesis is the correspondingly stronger bound $O(1/\sqrt{X})$ on the $o(1)$ oscillating term. At this intermediate scale, in contrast to $\psi(X) = X + O\big( \sqrt{X} (\log{X})^2 \big)$, a logarithmic factor in addition to the square root is not required, as $\sum_{\rho} 1 / |\rho|^2 < \infty$ over the zeros.

Here finally is how to deduce the more usual form $\psi(X) \sim X$ of the prime number theorem from the refinement $S(X) := \sum_{p < X} \Lambda(n)/n = \log{X} -\gamma + o(1)$ of Mertens's theorem: Summing by parts, $\psi(X) = XS(X) - \int_1^X S(t) \, dt = X(\log{X} - \gamma + o(1)) - \big( \int_1^X \log{t} \, dt - \gamma X + o(X) \big) = X + o(X)$.