Hi, I really don't know how to prove that the following series converges..
$\displaystyle \lim_{x \rightarrow \infty} \sum_{p \leq x}(p \log(p))^{-1} $
Thanks so much!!
Everk
I guess you need some heavy machinery for this. The n'th prime number $\displaystyle p_n$ satisfies $\displaystyle p_n\geqslant n\log n$, because of the prime number theorem. Also (obviously) $\displaystyle p_n>n$ and so $\displaystyle \log p_n>\log n.$
Therefore $\displaystyle \frac1{p_n\log p_n}<\frac1{n(\log n)^2}$. But $\displaystyle \sum\frac1{n(\log n)^2}$ converges, by the integral test. Hence, by the comparison test, so does $\displaystyle \sum\frac1{p_n\log p_n}.$
In...
http://www.mathhelpforum.com/math-he...ers-84832.html
... it has been demonstrated that...
$\displaystyle \sum_{k=2}^{n} \pi(k) \sim \ln (\ln n)$ (1)
... where...
$\displaystyle \pi(k) =\begin{cases}\frac{1}{k} &\text{k prime}\\ 0 &\text{elsewhere}\end{cases}$ (2)
Now is...
$\displaystyle \ln (\ln n) \sim \int_{2}^{n} \frac{dx}{x\ \ln x} = \ln (\ln n)-\ln (\ln 2)$ (3)
... so that...
$\displaystyle \sum_{k=2}^{n} \frac{\pi(k)}{\ln k} \sim \int_{2}^{n} \frac{dx}{x\ \ln^{2} x} = \frac{1}{\ln 2}-\frac{1}{\ln n}$ (4)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$