# Sum of reciprocals of Prime Numbers

• Apr 21st 2009, 10:19 AM
Fulger85
Sum of reciprocals of Prime Numbers
Hello,

Could anyone please give me a proof using basic/elementary number theory and or calculus of the following:

The sum of the reciprocals of primes is approximately equal to log(log(x))

e.g 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ... ~ log(log(x))

• Apr 21st 2009, 10:49 AM
chisigma
Simple question: what is 'x' in your formula?(Itwasntme)...

Kind regards

$\chi$ $\sigma$
• Apr 21st 2009, 11:01 AM
Fulger85
the sum of all reciprocal P's that are smaller or equal to x.

That is the x

Thanks for pointing it out.
• Apr 21st 2009, 06:46 PM
chisigma
All right!... the procedure that follows is due to Leohard Euler and starts with the 'infinite product' that brings his name...

$\sum_{k=1}^{\infty} \frac{1}{k^{n}}= \prod_{p} \frac{1}{1-p^{-n}}$ (1)

... where with 'p' we indicate the sequence of prime numbers. For n=1 the (1) becomes...

$\sum_{k=1}^{\infty} \frac{1}{k}= \prod_{p} \frac{1}{1-p^{-1}}$ (2)

Taking logarithm on both sides of (2)...

$\ln\sum_{k=1}^{\infty} \frac{1}{k}= \ln\prod_{p} \frac{1}{1-p^{-1}}= - \sum_{p} \ln (1-p^{-1}) = \sum_{p} (\frac{1}{p} + \frac{1}{2p^{2}} + \frac{1}{3p^{3}} + \dots) =$

$= \sum_{p} \frac{1}{p} + \sum_{p} \frac{1}{p^{2}}\cdot (\frac{1}{2} + \frac{1}{3p} + \frac{1}{4p^{2}}+ \dots) < \sum_{p} \frac{1}{p} + \sum_{p} \frac{1}{p^{2}}\cdot ( 1 + \frac{1}{p} + \frac{1}{p^{2}}+ \dots)=$

$= \sum_{p} \frac{1}{p} + \sum_{p} \frac{1}{p(p-1)} = \sum_{p} \frac{1}{p} + \chi = \sum_{k=1}^{\infty} \pi(k) + \chi$ (3)

... where $\chi$ is a positive constant and is $\pi(k)= \frac{1}{k}$ for k prime and $\pi(k)=0$ elsewhere. Euler had demonstrated before that...

$\lim_{n\rightarrow \infty} (\sum_{k=1}^{n}\frac{1}{k}- \ln n) = \gamma$ (4)

... where $\gamma$ is another positive constant and is $0<\gamma<1$, so that for n 'large enough' is ...

$\sum_{k=1}^{n}\frac{1}{k} \approx \ln n + \gamma$ (5)

... i.e. $\sum_{k=1}^{n}\frac{1}{k}$ is asymptotic to $\ln n$. Now observing (3) we conclude that $\sum_{p} \frac{1}{p}$ is asymptotic to $\ln(\ln n)$ i.e. for n 'large enough' is...

$\ln \sum_{k=1}^{n}\frac{1}{k} \approx \sum_{k=1}^{n}\pi(k) \approx \ln(\ln n)$ (6)

... or more precisely is...

$\lim_{n\rightarrow \infty} \frac{\sum_{k=1}^{n} \pi(k)}{\ln(\ln n)}=1$ (7)

Kind regards

$\chi$ $\sigma$
• Apr 21st 2009, 08:03 PM
Fulger85
Thank you :)
• Apr 22nd 2009, 10:40 AM
chisigma
In order to well undestand please observe this figure...

http://digilander.libero.it/luposabatini/MHF9.bmp

... where are reported...

a) in blue the function...

$\sum_{k=1}^{n} \pi(k)$

... where...

$\pi(k)= \frac{1}{k}$, k prime, $0$ elsewhere...

b) in red the function...

$\ln\sum_{k=1}^{n} \frac{1}{k}$

c) in grey the function...

$\ln (\ln n)$

What Euler has demonstrated two and half centuries ago [and that i replied now...] is that...

$\lim_{n\rightarrow \infty} \frac{\sum_{k=1}^{\infty} \pi(k)}{\ln (\ln n)}=1$

... and nothing else...

Kind regards

$\chi$ $\sigma$