Showing $\sum_{p^k\le x}{k\ge 2} = O(1/p^2)$

Question

I am trying to show that for primes $p$, $$\sum_{p\le x} \log p \sum_{p^k \le x, k\ge 2} \frac{1}{p^k} \ll \sum_{p\le x}\frac{\log p}{p^2}$$ for any $x\ge 2$.

I think I need to use geometric sum but that just gives me $\sum_{p^k\le x,k\ge 2}\frac{1}{p^k} < \frac{1}{p^2}\frac{p}{p-1}$ so I can't get a bound for all $x\ge 2$. How can I show this inequality?

Answer 1

Just use the properties of geometric series:

$$ \sum_{k\ge2}{1\over p^k}={1\over p^2}\sum_{k\ge2}{1\over p^{k-2}}={1\over p^2(1-p^{-1})}\le{2\over p^2}, $$

in which the last inequality follows from $p\ge2$.