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For $2\le z\le y$, some absolute $c>0$ gives

$$\sum_{z\le p\le y}\frac1p \ll\frac{e^{-c\sqrt{\log z}}+\log(y/z)}{\log z}. \tag{1}$$

Proof. Partial summation of the PNT in notation gives the more precise Mertens form

$$\sum_{p\le t}\frac1p =\log\log t+B+O(1/\log t)\qquad(t\ge10). \tag{2}$$

Indeed, the prime sum equals $\pi(t)/t+\int_2^t\pi(u)u^{-2}du$. Substituting $\pi(u)=\operatorname{li}(u)+E(u)$ makes the main part $\log\log t$ plus a constant. The integral of $E(u)/u^2$ converges, and its tail, together with $E(t)/t$, is $O(1/\log t)$ by the exponential PNT error. Bounded smaller $t$ can be absorbed.

If $y>2z$, (2), with the possible endpoint term $1/z$, gives

$$\sum_{z\le p\le y}\frac1p \le\log\log y-\log\log z+O(1/\log z) \ll\frac{\log(y/z)}{\log z}.$$

Here $\log(1+u)\le u$ and $\log(y/z)>\log2$ absorb the error. If $z\le y\le2z$ and $z\ge10$, Lemma 1.6 gives

$$\sum_{z\le p\le y}\frac1p \le\frac{\pi([z,y])}{z} \ll\frac{y-z}{z\log z}+e^{-c_0\sqrt{\log z}}.$$

On this range $(y-z)/z\ll\log(y/z)$. Decrease $c_0$ to $c>0$ so $e^{-c_0\sqrt{\log z}}\ll e^{-c\sqrt{\log z}}/\log z$. For $2\le z<10$ and $y\le2z$, the left side is bounded, while the exponential term divided by $\log z$ has a positive lower bound on that compact range. This completes every case. $\square$

Source. Tao, published paper, published p.799, Lemma 1.7. This page uses that published version.

Bears on. Problem 49.