I am reading the article *D. M. Gordon and C. Pomerance, The distribution of Lucas and elliptic pseudoprime, Math. Comp. (1991)* (click).
In equation (27) the authors, apparently, used the following upper bound for a sum over the prime numbers:
$$(\star)\quad\sum_{p \leq x} p^{-1+\varepsilon} \ll \frac{x^{\varepsilon}}{\varepsilon\log x} ,$$
for sufficiently large $x > 0$ and sufficiently small $\varepsilon > 0$, where the implicit constant in $\ll$ is absolute. (Precisely, $\varepsilon = \frac{4+\log\log\log x}{\log\log x}$, but I hope that this is not really important.)
I am trying to prove ($\star$) but since now I have failed. Clearly, one way can be using partial summation and the prime number theorem, for example
$$\sum_{p \leq x} p^{-1+\varepsilon} = \pi(x) x^{-1+\varepsilon} + (1-\varepsilon)\int_2^x \pi(t)t^{-2+\varepsilon}dt ,$$
but then I am now been able to prove the claim. I also throught that
$$\sum_{p \leq x} p^{-1+\varepsilon} \leq \sum_{n \leq \pi(x)} n^{-1+\varepsilon} \ll \int_0^{\pi(x)} t^{-1+\varepsilon}dt = \frac{\pi(x)^\varepsilon}{\varepsilon} \ll \frac{x^\varepsilon}{\varepsilon (\log x)^\varepsilon} ,$$
but his is too weak.

Thank you in advance for any suggestion.