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Source and scope. Compilation details for the analytic estimates in Croot's published paper, Lemma 2 on p. 235 and the expansion of (2) on pp. 233–234. These elementary estimates are proved here; no prime number theorem is required.

Statement. For $z\ge3$,

$$A(z)=\sum_{p\le z}\frac1p=O(\log\log z).$$

For $3/4\le\sigma<1$ and $y\ge3$, the finite Euler product satisfies

$$Z_y(\sigma)=\prod_{p\le y}(1-p^{-\sigma})^{-1}, \qquad \log Z_y(\sigma)\le y^{1-\sigma}A(y)+O(1),$$

with an absolute constant in the last error term.

Complete proof. Set $s=1+1/\log z$. For $p\le z$, $p^{s-1}\le e$, so

$$A(z)\le e\sum_p p^{-s}\le e\log\zeta(s).$$

The last inequality follows by expanding the absolutely convergent Euler product for $\zeta(s)$ and retaining the first power of each prime. The integral comparison

$$\zeta(s)=\sum_{n\ge1}n^{-s} \le1+\int_1^\infty t^{-s}\,dt=1+\frac1{s-1}$$

therefore gives $A(z)\le e\log(1+\log z)$.

For the second assertion, the first powers in the logarithmic expansion obey

$$\sum_{p\le y}p^{-\sigma} =\sum_{p\le y}\frac{p^{1-\sigma}}p \le y^{1-\sigma}A(y).$$

The remaining powers are bounded uniformly:

$$\sum_{p\le y}\sum_{a\ge2}\frac{p^{-a\sigma}}a \le\sum_{p\le y}\frac{p^{-2\sigma}}{1-p^{-\sigma}} \le\frac1{1-2^{-3/4}}\sum_{m\ge2}m^{-3/2}<\infty.$$

Adding the two estimates proves the result. The Euler-product identities used here follow by expanding finite products and then taking monotone limits of nonnegative absolutely convergent series.

Bears on. Lemma 2 and prime-power smoothness.