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For $y\ge10$ and nonnegative real numbers $a_d$ indexed by $d\in\mathbb N_{\le y}$,

$$\sum_{d\in\mathbb N_{\le y}}a_d \ll(\log y)\sup_{d\in\mathbb N_{\le y}} d^{1-1/\log y}a_d. \tag{1}$$

Either side may be infinite. The notation and exact classical inputs are fixed in Definitions and analytic inputs.

Proof. Set $\alpha=1-1/\log y>0$. Unique factorization and monotone limits of positive geometric series give

$$\sum_{d\in\mathbb N_{\le y}}d^{-\alpha} =\prod_{p\le y}(1-p^{-\alpha})^{-1}.$$

For $y\ge e^4$, $\alpha\ge3/4$, so uniformly in $p\le y$,

$$(1-p^{-\alpha})^{-1} =1+\frac{p^{1/\log y}}p+O(p^{-3/2}).$$

Since $0\le\log p/\log y\le1$, the inequality $e^u=1+O(u)$ on $[0,1]$ yields

$$\log\prod_{p\le y}(1-p^{-\alpha})^{-1} =\sum_{p\le y}\frac1p+ O\!\left(\frac1{\log y}\sum_{p\le y}\frac{\log p}{p}\right)+O(1) =\log\log y+O(1).$$

The error $\sum p^{-3/2}$ converges, and the two prime sums are (2) in the input page. For $10\le y\le e^4$, only finitely many primes occur and $\alpha\ge1-1/\log10>0$. The product is uniformly bounded; enlarging the constant therefore proves the same $O(\log y)$ bound.

Let $T$ be the supremum in (1). If $T=\infty$, the assertion is immediate. Otherwise $a_d\le T d^{-\alpha}$ for every $d$, and summing proves (1). This includes $T=0$ and the term $d=1$. $\square$

Source. Tao, published paper, published pp.798–799, Lemma 1.5. This page uses that published version.

Bears on. Problem 49.