Showing $\sum_{j \geq 2}\sum_{x^{1/j} \leq p \leq x}\frac{1}{jp^{j}} = O\left(\frac{1}{\log x}\right)$

Let $p$ denote a prime. Suppose I am given the asymptotic that $$\sum_{1 \leq n \leq x} \frac{\Lambda(n)}{n\log n} = \log\log x + \gamma + O\left(\frac{1}{\log x}\right),$$ why is $$\sum_{2 \leq p \leq x}\sum_{j = 1}^{\infty}\frac{1}{jp^{j}} = \log\log x + \gamma + O\left(\frac{1}{\log x}\right)?$$

To show this we need to show that $$\sum_{j \geq 2}\sum_{x^{1/j} \leq p \leq x}\frac{1}{jp^{j}} = O\left(\frac{1}{\log x}\right).$$ Are there any suggestions on how to go about doing this?

• $\Lambda$ appears to be the Mangoldt function, and $p$ runs through prime numbers. Jan 20, 2015 at 0:47
• It should probably be $p > x^{1/j}$. Jan 20, 2015 at 0:59

Ignoring the fact that $p$ is required to be prime, for a fixed $j$ you can establish the inequality $$\sum_{n > x^{1/j}} \frac{1}{jn^j} \leq \frac{1}{j(j-1)}x^{1/j-1} + 1/jx$$ by comparison with $\displaystyle \int_{x^{1/j}}^{+\infty} \frac{dt}{t^j}$.
Summing over $j$, noting that $\sum 1/j(j-1) = 1$, and taking into account that at most $(\log x)/(\log 2)$ vales of $j$ can appear in the sum (before there are no prime numbers left in it), we find the estimate
$$\sum_{j \geq 2} \sum_{n > x^{1/j}} \frac{1}{jn^j} \leq \frac{1}{\sqrt{x}} + \frac{1}{2\log 2}\frac{\log x}{x}.$$