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KConrad
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The Wikipedia page you mention no longer refers to an equivalence between PNT and the estimate on $\sum_{p \leq x} 1/p$ with error term $o(1/\log x)$: the View History tab of that page shows an edit was made to iton Jan. 10, 2023 by RStanley31 (gee, who could that be... :)) with the comment "Result of equivalence is not found within the literature. The forward implication is found within Tenebaum’s book as mention in the reference."

In another answer here, GH from MO shows the forward implication: PNT implies $\sum_{n \leq x} 1/p = \log x + C + o(1)$ for asome constant $C$ (a hard step), and that implies $\sum_{p \leq x} 1/p = \log\log x + M + o(1/\log x)$ for some constant $M$ by partial summation. Here is a converse argument, from the estimate on $\sum_{p\leq x}1/p$ with error term $o(1/\log x)$ to PNT, thus establishing that there is an equivalence $$ {\rm PNT} \Longleftrightarrow \sum_{p \leq x} \frac{1}{p} = \log\log x + M + o\left(\frac{1}{\log x}\right) $$ for some constant $M$.

Set $A(x) = \sum_{p \leq x} 1/p$, so we assume $A(x) = \log\log x + M + o(1/\log x)$ for some $M$, as $x \to \infty$. Then $$ \pi(x) = \sum_{p \leq x} 1 = \sum_{p \leq x} \frac{1}{p}p = \sum_{n \leq x} a_n n $$ where $a_n = 1/p$ when $n = p$ is prime and $a_n = 0$ otherwise. Then $A(x) = \sum_{n \leq x} a_n$, so by partial summation $$ \sum_{n \leq x} a_n n = A(x)x - \int_2^x A(y)\,dy $$ for $x \geq 2$. Since $A(x) = \log\log x + M + o(1/\log x)$, $$ A(x)x = x\log\log x + Mx + o\left(\frac{x}{\log x}\right) $$ and $$ \int_2^x A(y)\,dy = \int_2^x \log\log y\,dy + M(x-2) + \int_2^x o\left(\frac{1}{\log y}\right)\,dy. $$ Using integration by parts, $$ \int_2^x \log\log y\,dy = x\log\log x - 2\log\log 2 - \int_2^x\frac{dy}{\log y}. $$ Putting these formulas into the expression for $\pi(x)$ as $\sum_{n \leq x} a_n n$, we get $$ \pi(x) = \int_2^x \frac{dy}{\log y} + 2M + 2\log\log 2 + o\left(\frac{x}{\log x}\right) - \int_2^x o\left(\frac{1}{\log y}\right)\,dy. $$ The constant $2M + 2\log\log 2$ can be absorbed into the $o(x/\log x)$ term.

Lastly, since $\int_2^x o(1/\log y)\,dy = o(\int_2^x dy/\log y)$ as $x \to \infty$, we have $$ \pi(x) = \int_2^x \frac{dy}{\log y} + o\left(\frac{x}{\log x}\right), $$ which is PNT.

PS. The Wikipedia page you mention no longer refers to an equivalence between PNT and the estimate on $\sum_{p \leq x} 1/p$ with error term $o(1/\log x)$, but only a one-way implication from PNT to the estimate on $\sum_{p \leq x} 1/p$: the View History tab of that page shows an edit was made to it on Jan. 10, 2023 by RStanley31 (gee, who could that be... :)) with the comment "Result of equivalence is not found within the literature." Perhaps that page should be reverted to the original wording with a link to this page.

The Wikipedia page you mention no longer refers to an equivalence between PNT and the estimate on $\sum_{p \leq x} 1/p$ with error term $o(1/\log x)$: the View History tab of that page shows an edit was made to iton Jan. 10, 2023 by RStanley31 (gee, who could that be... :)) with the comment "Result of equivalence is not found within the literature. The forward implication is found within Tenebaum’s book as mention in the reference."

In another answer here, GH from MO shows the forward implication: PNT implies $\sum_{n \leq x} 1/p = \log x + C + o(1)$ for a constant $C$ (a hard step), and that implies $\sum_{p \leq x} 1/p = \log\log x + M + o(1/\log x)$ for some constant $M$ by partial summation. Here is a converse argument, from the estimate on $\sum_{p\leq x}1/p$ with error term $o(1/\log x)$ to PNT, thus establishing that there is an equivalence $$ {\rm PNT} \Longleftrightarrow \sum_{p \leq x} \frac{1}{p} = \log\log x + M + o\left(\frac{1}{\log x}\right) $$ for some constant $M$.

Set $A(x) = \sum_{p \leq x} 1/p$, so we assume $A(x) = \log\log x + M + o(1/\log x)$ for some $M$, as $x \to \infty$. Then $$ \pi(x) = \sum_{p \leq x} 1 = \sum_{p \leq x} \frac{1}{p}p = \sum_{n \leq x} a_n n $$ where $a_n = 1/p$ when $n = p$ is prime and $a_n = 0$ otherwise. Then $A(x) = \sum_{n \leq x} a_n$, so by partial summation $$ \sum_{n \leq x} a_n n = A(x)x - \int_2^x A(y)\,dy $$ for $x \geq 2$. Since $A(x) = \log\log x + M + o(1/\log x)$, $$ A(x)x = x\log\log x + Mx + o\left(\frac{x}{\log x}\right) $$ and $$ \int_2^x A(y)\,dy = \int_2^x \log\log y\,dy + M(x-2) + \int_2^x o\left(\frac{1}{\log y}\right)\,dy. $$ Using integration by parts, $$ \int_2^x \log\log y\,dy = x\log\log x - 2\log\log 2 - \int_2^x\frac{dy}{\log y}. $$ Putting these formulas into the expression for $\pi(x)$ as $\sum_{n \leq x} a_n n$, we get $$ \pi(x) = \int_2^x \frac{dy}{\log y} + 2M + 2\log\log 2 + o\left(\frac{x}{\log x}\right) - \int_2^x o\left(\frac{1}{\log y}\right)\,dy. $$ The constant $2M + 2\log\log 2$ can be absorbed into the $o(x/\log x)$ term.

Lastly, since $\int_2^x o(1/\log y)\,dy = o(\int_2^x dy/\log y)$ as $x \to \infty$, we have $$ \pi(x) = \int_2^x \frac{dy}{\log y} + o\left(\frac{x}{\log x}\right), $$ which is PNT.

In another answer here, GH from MO shows PNT implies $\sum_{n \leq x} 1/p = \log x + C + o(1)$ for some constant $C$ (a hard step), and that implies $\sum_{p \leq x} 1/p = \log\log x + M + o(1/\log x)$ for some constant $M$ by partial summation. Here is a converse argument, from the estimate on $\sum_{p\leq x}1/p$ with error term $o(1/\log x)$ to PNT, thus establishing that there is an equivalence $$ {\rm PNT} \Longleftrightarrow \sum_{p \leq x} \frac{1}{p} = \log\log x + M + o\left(\frac{1}{\log x}\right) $$ for some constant $M$.

Set $A(x) = \sum_{p \leq x} 1/p$, so we assume $A(x) = \log\log x + M + o(1/\log x)$ for some $M$, as $x \to \infty$. Then $$ \pi(x) = \sum_{p \leq x} 1 = \sum_{p \leq x} \frac{1}{p}p = \sum_{n \leq x} a_n n $$ where $a_n = 1/p$ when $n = p$ is prime and $a_n = 0$ otherwise. Then $A(x) = \sum_{n \leq x} a_n$, so by partial summation $$ \sum_{n \leq x} a_n n = A(x)x - \int_2^x A(y)\,dy $$ for $x \geq 2$. Since $A(x) = \log\log x + M + o(1/\log x)$, $$ A(x)x = x\log\log x + Mx + o\left(\frac{x}{\log x}\right) $$ and $$ \int_2^x A(y)\,dy = \int_2^x \log\log y\,dy + M(x-2) + \int_2^x o\left(\frac{1}{\log y}\right)\,dy. $$ Using integration by parts, $$ \int_2^x \log\log y\,dy = x\log\log x - 2\log\log 2 - \int_2^x\frac{dy}{\log y}. $$ Putting these formulas into the expression for $\pi(x)$ as $\sum_{n \leq x} a_n n$, we get $$ \pi(x) = \int_2^x \frac{dy}{\log y} + 2M + 2\log\log 2 + o\left(\frac{x}{\log x}\right) - \int_2^x o\left(\frac{1}{\log y}\right)\,dy. $$ The constant $2M + 2\log\log 2$ can be absorbed into the $o(x/\log x)$ term.

Lastly, since $\int_2^x o(1/\log y)\,dy = o(\int_2^x dy/\log y)$ as $x \to \infty$, we have $$ \pi(x) = \int_2^x \frac{dy}{\log y} + o\left(\frac{x}{\log x}\right), $$ which is PNT.

PS. The Wikipedia page you mention no longer refers to an equivalence between PNT and the estimate on $\sum_{p \leq x} 1/p$ with error term $o(1/\log x)$, but only a one-way implication from PNT to the estimate on $\sum_{p \leq x} 1/p$: the View History tab of that page shows an edit was made to it on Jan. 10, 2023 by RStanley31 (gee, who could that be... :)) with the comment "Result of equivalence is not found within the literature." Perhaps that page should be reverted to the original wording with a link to this page.

Source Link
KConrad
KConrad
  • 50.6k
  • 9
  • 196
  • 277

The Wikipedia page you mention no longer refers to an equivalence between PNT and the estimate on $\sum_{p \leq x} 1/p$ with error term $o(1/\log x)$: the View History tab of that page shows an edit was made to iton Jan. 10, 2023 by RStanley31 (gee, who could that be... :)) with the comment "Result of equivalence is not found within the literature. The forward implication is found within Tenebaum’s book as mention in the reference."

In another answer here, GH from MO shows the forward implication: PNT implies $\sum_{n \leq x} 1/p = \log x + C + o(1)$ for a constant $C$ (a hard step), and that implies $\sum_{p \leq x} 1/p = \log\log x + M + o(1/\log x)$ for some constant $M$ by partial summation. Here is a converse argument, from the estimate on $\sum_{p\leq x}1/p$ with error term $o(1/\log x)$ to PNT, thus establishing that there is an equivalence $$ {\rm PNT} \Longleftrightarrow \sum_{p \leq x} \frac{1}{p} = \log\log x + M + o\left(\frac{1}{\log x}\right) $$ for some constant $M$.

Set $A(x) = \sum_{p \leq x} 1/p$, so we assume $A(x) = \log\log x + M + o(1/\log x)$ for some $M$, as $x \to \infty$. Then $$ \pi(x) = \sum_{p \leq x} 1 = \sum_{p \leq x} \frac{1}{p}p = \sum_{n \leq x} a_n n $$ where $a_n = 1/p$ when $n = p$ is prime and $a_n = 0$ otherwise. Then $A(x) = \sum_{n \leq x} a_n$, so by partial summation $$ \sum_{n \leq x} a_n n = A(x)x - \int_2^x A(y)\,dy $$ for $x \geq 2$. Since $A(x) = \log\log x + M + o(1/\log x)$, $$ A(x)x = x\log\log x + Mx + o\left(\frac{x}{\log x}\right) $$ and $$ \int_2^x A(y)\,dy = \int_2^x \log\log y\,dy + M(x-2) + \int_2^x o\left(\frac{1}{\log y}\right)\,dy. $$ Using integration by parts, $$ \int_2^x \log\log y\,dy = x\log\log x - 2\log\log 2 - \int_2^x\frac{dy}{\log y}. $$ Putting these formulas into the expression for $\pi(x)$ as $\sum_{n \leq x} a_n n$, we get $$ \pi(x) = \int_2^x \frac{dy}{\log y} + 2M + 2\log\log 2 + o\left(\frac{x}{\log x}\right) - \int_2^x o\left(\frac{1}{\log y}\right)\,dy. $$ The constant $2M + 2\log\log 2$ can be absorbed into the $o(x/\log x)$ term.

Lastly, since $\int_2^x o(1/\log y)\,dy = o(\int_2^x dy/\log y)$ as $x \to \infty$, we have $$ \pi(x) = \int_2^x \frac{dy}{\log y} + o\left(\frac{x}{\log x}\right), $$ which is PNT.