Let $p$ be a prime. Then it has been proven by Erdős that $$\sum_{p\mid 2^{n}-1}\frac1{p} \ll \log \log \log n.$$ Let $c<1$ be a positive constant. Then can we prove that $$\sum_{p\mid 2^{n}-1}\frac1{p^{c}}\ll \frac{\log ^{1-c} n}{\log \log n}?$$

It seems to me that arguments of Erdős should apply to this case and one should be able to obtain this bound. Thanks in advance for any assistance.

Let $p$ be a prime. Then it has been proven by Erdős that $$\sum_{p\mid 2^{n}-1}\frac1{p} \ll \log \log \log n.$$ Let $c<1$ be a positive constant. Then can we prove that $$\sum_{p\mid 2^{n}-1}\frac1{p^{c}}\ll \frac{\log ^{1-c} n}{\log \log n}?$$

Thanks in advance for any assistance.