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

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- Sep 4th 2011, 06:14 PMeverkSeries Involving Prime Numbers
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 - Sep 4th 2011, 07:04 PMILoveMaths07Re: Series Involving Prime Numbers
Use the integral test.

- Sep 5th 2011, 08:27 AMeverkRe: Series Involving Prime Numbers
- Sep 5th 2011, 12:45 PMOpalgRe: Series Involving Prime Numbers
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 $\displaystyle \sum\frac1{p_n\log p_n}.$ - Sep 5th 2011, 01:14 PMchisigmaRe: Series Involving Prime Numbers
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

$\displaystyle \chi$ $\displaystyle \sigma$