2
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ $\Lambda$ appears to be the Mangoldt function, and $p$ runs through prime numbers. $\endgroup$
    – user208259
    Jan 20, 2015 at 0:47
  • $\begingroup$ It should probably be $p > x^{1/j}$. $\endgroup$
    – user208259
    Jan 20, 2015 at 0:59

1 Answer 1

1
$\begingroup$

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

$$\sum_{j \geq 2} \sum_{n > x^{1/j}} \frac{1}{jn^j} \leq \frac{1}{\sqrt{x}} + \frac{1}{2\log 2}\frac{\log x}{x}.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .