Mertens-like theorem

Question

Mertens' first theorem states that $$ \sum_{p \leq n} \frac{\log p}{p} = \log n + O(1). $$ I read in this paper that the following variant is "classical": $$ \sum_{p \leq n} \frac{\log p}{p - 1} = \log n - \gamma + o(1). $$ Could anyone provide a reference (or a proof) to the "classical" variant?

Answer 1

This lies beyond Mertens, in the sense that this variant actually implies the Prime Number Theorem, as will be explained below, while Mertens' theorem is weaker than the PNT.

I sketch below a complex analytic proof of the variant, due to Landau.

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Let $\Lambda$ be the von Mangoldt function, defined as $\Lambda(p^k)=\log p$ if $p$ is a prime and $k \ge 1$, and $\Lambda(n)=0$ otherwise. Note that your sum $$S_1(n):=\sum_{p \le n} \frac{\log p}{p-1}$$ is very close to $$S_2(n):= \sum_{m \le n} \frac{\Lambda(m)}{m}.$$ Indeed, $$S_2(n)-S_1(n) = \sum_{p \le n} \log p \sum_{1 \le k \le \log_p n} p^{-k}- \sum_{p \le n} \log p\sum_{k \ge 1} p^{-k} = \sum_{p \le n} \log p \sum_{k >\log _p n} p^{-k}.$$ The inner $k$-sum is always $\ll \min\{1/n, 1/p^2\}$, so that this difference is
$$\ll \sum_{p \le n} \log p \min\{1/n, 1/p^2\}\ll \sum_{m \le \sqrt{n}} \frac{\log m}{n} + \sum_{\sqrt{n}<m \le n} \frac{\log m}{m^2} \ll \frac{\log n}{\sqrt{n}}=o(1).$$ We see it suffices to estimate $S_2(n)$.

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The reason that I introduced $\Lambda$ is that the PNT is usually proved with $\Lambda$. The Dirichlert series of $\Lambda$ is $-\zeta'(s)/\zeta(s)$, so Perron's formula gives $$S_2(n) = \frac{1}{2\pi i}\int_{(1)}-\frac{\zeta'}{\zeta}(s+1) n^s \frac{ds}{s}+O((\log n)/n).$$ The main term $\log n - \gamma$ arises as the residue of the double pole of the integrand at $s=0$. Indeed, near $s=0$ we have the Laurent expansion $$\begin{align}-\frac{\zeta'}{\zeta}(s+1)n^s/s&= s^{-2} (1-\gamma s + O(s^2)) (1+s \log n + O(s^2 \log^2 n)) \\ &= s^{-2}(1+(\log n-\gamma)s+O(s^2)). \end{align}$$ Here we made use of $$(\star)\, -\frac{\zeta'}{\zeta}(s) = \frac{1}{s-1} -\gamma + O(s-1)$$ near $s=1$. To justify the above heuristic one needs, as in the classical proof of the PNT, to 1) apply a truncated version of Perron's formula and 2) shift the contour to the left using a zero-free region. This is why I believe Hadamard and de la Vallée Poussin had possibly known this variant.

Landau, in his book "Handbuch der Lehre von der Verteilung der Primzahlen" (1911) used Perron's formula to work out an explicit formula for $\sum_{m \le n} \Lambda(m)/m^s$ for general $s$. For $s=1$ it corresponds to $S_2(n)$. A reference with proof is Lemma 4.1 in this preprint (specialize to $s=1$). In particular, letting $$\Delta(n):=S_2(n)-( \log n- \gamma)$$ then, assuming $n$ is a positive integer, $$\Delta(n)=\frac{\Lambda(n)}{2n}- \sum_{\rho} \frac{n^{\rho-1}}{\rho-1} \ll \exp(-c(\log n)^{3/5}(\log \log n)^{-1/5})$$ for some $c>0$. Here the sum is over the zeros of $\zeta$, and the inequality follows from the Vinogradov--Korobov zero-free region. Under RH, $\Delta(n)\ll n^{-1/2}\log^2 n$.

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The Laurent expansion $(\star)$ is not mysterious; it's equivalent to the Laurent expansion $$(\star \star)\, \zeta(s) = \frac{1}{s-1} + \gamma +O(s-1)$$ at the same point, which in turn is equivalent to the Harmonic sum $\sum_{m \le n} 1/n$ being equal to $\log n + \gamma + O(1/n)$. See Corollary 1.16 in Montgomery and Vaughan's book, "Multiplicative Number Theory I" for a rigorous derivation of $(\star \star)$.

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Here is a more accessible proof. Given an arithmetic function $f=\alpha*\beta$ we have $$\sum_{n \le x} f(n)=\sum_{n \le x} \alpha(n) \sum_{m\le x/n} \beta(m).$$ Applying this with $\log = \mathbf{1}*\Lambda$ where $\log$ is the natural logarithm and $\mathbf{1}$ is the constant function taking the value $1$, we find $$\sum_{n \le x} \log n = \sum_{n \le x} \Lambda(n)\lfloor x/n \rfloor.$$ This identity goes back to Chebyshev. Dividing both sides by $x$ we see $$\sum_{n \le x} \frac{\Lambda(n)}{n} =x^{-1}\sum_{n \le x} \log n +x^{-1} \sum_{n \le x} \Lambda(n) \{x/n\}$$ where $\{t\} \in [0,1)$ is the fractional part of $t$. A very weak version of Stirling's approximation tells us $$x^{-1}\sum_{n \le x} \log n=\log x - 1 + O\left( \frac{\log x}{x}\right).$$ To establish $S_2(x) =\log x- \gamma+o(1)$ it remains to show $$x^{-1} \sum_{n \le x} \Lambda(n) \{x/n\} = 1-\gamma + o(1).$$ By the Prime Number Theorem, $\Lambda =1$ 'on average', so we would expect the sum to be $$=x^{-1} \sum_{n \le x} \{x/n\}+o(1)$$ which indeed tends to $1-\gamma$ (see Exercise 2.1.1.1, page 39 in Montgomery--Vaughan, which is attributed to de la Vallée Poussin; this is a restatement of the asymptotic $\sum_{n \le x} d(n) = x\log x + (2\gamma-1)x+o(x)$ which was known to Dirichlet). To make this formal we just need to show that the difference $$\sum_{n \le x} (\Lambda(n)-1) \{x/n\}$$ is $o(x)$ which can be done via the Prime Number Theorem, see Exercise 8.1.1.1, page 248, in Montgomery--Vaughan. In the same exercise they use Axer's theorem to show that $\sum_{n\le x}\Lambda(n) \sim x$ (PNT) is in fact equivalent to $S_2(n) =\log n- \gamma+o(1)$.

Answer 2

I'll use the same notation $S_1(x)$ and $S_2(x)$ as in Ofir's answer and give more details behind some of the estimates and rely on a different approach to the Prime Number Theorem (no use of zero-free regions for $\zeta(s)$ to the left of the line ${\rm Re}(s) = 1$).

Set $$ S_1(x) = \sum_{p \leq x} \frac{\log p}{p-1} = \sum_{p \leq x} \log p\frac{1/p}{1-1/p} = \sum_{p \leq x} (\log p)\sum_{k \geq 1} \frac{1}{p^k} = \sum_{p \leq x} \sum_{k \geq 1} \frac{\log p}{p^k} $$ and $$ S_2(x) = \sum_{p^k \leq x} \frac{\log p}{p^k}. $$ We intend $k \geq 1$ here, so $S_2(x)$ only involves primes $p \leq x$ and, for each $p$, positive integers $k \leq \log_p x$.

Each term in $S_2(x)$ is part of the series $S_1(x)$, so let's extract it. Since $p^k \leq x$ is the same as $k \leq \log_p x$, split the inner sum in $S_1(x)$ according to a comparison between $k$ and $\log_p x$: $$ S_1(x) = \sum_{p \leq x} \left(\sum_{k \leq \log_p x}\frac{\log p}{p^k} + \sum_{k > \log _p x} \frac{\log p}{p^k}\right). $$

Breaking off the first (finite) inner sum from the rest, $$ S_1(x) = \sum_{p \leq x} \sum_{k \leq \log_p x} \frac{\log p}{p^k} + \sum_{p \leq x} \sum_{k > \log_p x} \frac{\log p}{p^k} = S_2(x) + \sum_{p \leq x} \sum_{k > \log_p x} \frac{\log p}{p^k}, $$ so $$ 0 \leq S_1(x) - S_2(x) = \sum_{p \leq x} (\log p)\sum_{k > \log_p x} \frac{1}{p^k}. $$

This double sum for $S_1(x) - S_2(x)$ has no inner term with $k = 1$: if $k > \log_p x$ then $p^k > x \geq p$, so $k \geq 2$. We'll bound the inner sum in two ways.

First we bound $\sum_{k > y} 1/p^k$ for all positive real $y$. Let $N$ be the positive integer such that $y < N \leq y+1$, so $$ \sum_{k > y} \frac{1}{p^k} = \sum_{k \geq N} \frac{1}{p^k} = \frac{1}{p^N}\frac{1}{1-1/p} < \frac{1}{p^y} \frac{1}{1-1/p} \leq \frac{2}{p^y} $$ since $p \geq 2$. Taking $y = \log_p x$, we get $$ \sum_{k > \log_p x} \frac{1}{p^k} < \frac{2}{x}. $$

Since all terms in the inner sum for $S_1(x) - S_2(x)$ have $k \geq 2$, $$ \sum_{k > \log_p x} \frac{1}{p^k} \leq \sum_{k \geq 2} \frac{1}{p^k} = \frac{1/p^2}{1-1/p} \leq \frac{2}{p^2}. $$ So the inner sum in $S_1(x)-S_2(x)$ is bounded above by both $2/x$ and $2/p^2$. We want to use the smaller of these.

For $p \leq x$, we have $1/x \leq 1/p^2$ when $p \leq \sqrt{x}$, and we have $1/p^2 < 1/x$ when $x < p^2$, or equivalently $\sqrt{x} < p \leq x$. So let's split up the outer sum condition $p \leq x$ for $S_1(x) - S_2(x)$ into $p \leq \sqrt{x}$ and $\sqrt{x} < p \leq x$: $$ 0 \leq S_1(x) - S_2(x) = \sum_{p \leq \sqrt{x}} (\log p)\sum_{k > \log_p x} \frac{1}{p^k} + \sum_{\sqrt{x} < p \leq x} (\log p)\sum_{k > \log_p x} \frac{1}{p^k}. $$ For the first inner sum use the upper bound $2/x$, and for the second inner sum use the upper bound $2/p^2$: $$ 0 \leq S_1(x) - S_2(x) \leq \sum_{p \leq \sqrt{x}} \frac{2\log p}{x} + \sum_{\sqrt{x} < p \leq x} \frac{2\log p}{p^2} = \frac{2}{x}\sum_{p \leq \sqrt{x}}\log p + \sum_{\sqrt{x} < m \leq x} \frac{2\log m}{m^2} $$ where the second sum runs over positive integers $m$.

Chebyshev showed $\sum_{p \leq y} \log p = O(y)$ as $y \to \infty$, so in the last expression above, the first term is $O(\sqrt{x}/x) = O(1/\sqrt{x})$. Since $(\log t)/t^2$ is decreasing for large $t$ (really, for $t \geq \sqrt{e}$), the second term grows like $\int_{\sqrt{x}}^x (\log t)/t^2\,dt$. An antiderivative of $(\log t)/t^2$ is $-(1+\log t)/t$, so the integral is $-1/x - (\log x)/x + 1/\sqrt{x} + (\log\sqrt{x})/\sqrt{x} = O((\log x)/\sqrt{x})$. Thus $S_1(x) - S_2(x)$ is $O((\log x)/\sqrt{x})$, so $S_1(x) - S_2(x) \to 0$ as $x \to \infty$. Hence showing $S_1(x) = \log x + c + o(1)$ for some constant $c$ is equivalent to showing $S_2(x) = \log x + c + o(1)$ for the same $c$.

It remains to explain why $$ \sum_{p^k \leq x} \frac{\log p}{p^k} = \log x - \gamma + o(1). $$

Remark. It turns out that showing $\sum_{p^k \leq x} (\log p)/p^k = \log x + c + o(1)$ for some constant $c$ is equivalent to the Prime Number Theorem. Because the subseries with $k \geq 2$ converges, that estimate is equivalent to saying $\sum_{p \leq x} (\log p)/p = \log x + c' + o(1)$ for some constant $c'$. That this estimate is equivalent to the Prime Number Theorem is discussed on an MO page here.

Newman's method of proving the Prime Number Theorem proves the following result.

Theorem. Let $a_n \geq 0$ for all $n$ and assume $f(s) = \sum_{n \geq 1} a_n/n^s$ converges for ${\rm Re}(s) > 1$. Assume also that $f(s)$ has an analytic continuation to the line ${\rm Re}(s) = 1$ except for at worse a simple pole at $s = 1$, where $f(s) = r/(s-1) + c_0 + O(s-1)$ for $s$ near $1$. If $\sum_{n \leq x} a_n = O(x)$ then $$ \frac{1}{x}\sum_{n \leq x} a_n \to r \ {\it and } \ \sum_{n \leq x} \frac{a_n}{n} = r\log x + c_0 + o(1) $$ as $x \to \infty$.

This theorem is applied to the negative logarithmic derivative $$ f(s) = -\frac{\zeta'(s)}{\zeta(s)} = \sum_{p^k} \frac{\log p}{p^{ks}}, $$ where we have that Dirichlet series over prime powers from the Euler product for $\zeta(s)$. The function $\zeta(s)$ extends analytically from ${\rm Re}(s) > 1$ to ${\rm Re}(s) \geq 1$ except for a simple pole at $s = 1$ with residue $1$, and $\zeta(s) \not= 0$ on ${\rm Re}(s) \geq 1$.

The constant term in the Laurent expansion of $\zeta(s)$ at $s = 1$ turns out to be $\gamma$: $\zeta(s) = 1/(s-1) + \gamma + O(s-1)$ for $s$ near $1$. Using the Laurent expansion, $\zeta'(s) = -1/(s-1)^2 + O(1)$, so $$ -\zeta'(s) = \frac{1}{(s-1)^2} + O(1). $$ From $$ \zeta(s) = \frac{1}{s-1}(1 + \gamma(s-1) + O((s-1)^2)) $$ near $s = 1$, we get $$ \frac{1}{\zeta(s)} = (s-1)(1 - \gamma(s-1) + O((s-1)^2)) = (s-1) - \gamma(s-1)^2 + O((s-1)^3), $$ and using these expansions for $-\zeta'(s)$ and $1/\zeta(s)$ near $s = 1$ yields $$ -\frac{\zeta'(s)}{\zeta(s)} = \frac{1}{s-1} - \gamma + O(s-1). $$ Therefore Newman's theorem with $f(s) = -\zeta'(s)/\zeta(s)$, $r = 1$, and $c_0 = -\gamma$ implies $$ \frac{1}{x}\sum_{p^k \leq x} \log p \to 1 \ {\rm and } \ \sum_{p^k \leq x} \frac{\log p}{p^k} = r_0\log x + c_0 + o(1)= \log x - \gamma + o(1) $$ as $x \to \infty$. The first relation is a standard equivalent statement to the Prime Number Theorem and the second relation is the desired estimate on $S_2(x)$.