Thread: Sum of Reciprocal of primes

1. Sum of Reciprocal of primes

I'm trying to prove that if s is real and s>1 then

I have managed to use the euler product for the zeta function and the taylor expansion log function on the LHS to give

|sigma {p} sigma {m=2 to infinity} (1/(mp^ms))| (sorry my latex is bad)

Does anyone know where I go from here

2. Originally Posted by klw289
I'm trying to prove that if s is real and s>1 then

I have managed to use the euler product for the zeta function and the taylor expansion log function on the LHS to give

|sigma {p} sigma {m=2 to infinity} (1/(mp^ms))| (sorry my latex is bad)

Does anyone know where I go from here
From the Euler's product...

$\zeta(s)= \prod_{p} \frac{1}{1-p^{-s}}$ (1)

... You derive with simple steps...

$\ln \zeta(s)= \sum_{p} p^{-s} + \frac{1}{2} \sum_{p} p^{-2s} + \frac{1}{3} \sum_{p} p^{-3s} + ...$ (2)

... so that is...

$\ln \zeta(s) - \sum_{p} p^{-s} = \frac{1}{2} \sum_{p} p^{-2s} + \frac{1}{3} \sum_{p} p^{-3s} + ...$ (3)

Now we can proceed expanding in series the ratio $\frac{\zeta(2s)}{1-2^{-s}}$ ... and that is left to You...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
From the Euler's product...

$\zeta(s)= \prod_{p} \frac{1}{1-p^{-s}}$ (1)

... You derive with simple steps...

$\ln \zeta(s)= \sum_{p} p^{-s} + \frac{1}{2} \sum_{p} p^{-2s} + \frac{1}{3} \sum_{p} p^{-3s} + ...$ (2)

... so that is...

$\ln \zeta(s) - \sum_{p} p^{-s} = \frac{1}{2} \sum_{p} p^{-2s} + \frac{1}{3} \sum_{p} p^{-3s} + ...$ (3)

Now we can proceed expanding in series the ratio $\frac{\zeta(2s)}{1-2^{-s}}$ ... and that is left to You...

Kind regards

$\chi$ $\sigma$
Unfortunatly as I wrote in my first post I had managed already to reach

$\ln \zeta(s) - \sum_{p} p^{-s} = \sum_{p} \sum_{m=2}\frac{1}{m} \ p^{-ms}$

That is where I am now confused I had tried expanding the series but I really couldn't see the link

4. Originally Posted by chisigma
From the Euler's product...

$\zeta(s)= \prod_{p} \frac{1}{1-p^{-s}}$ (1)

... You derive with simple steps...

$\ln \zeta(s)= \sum_{p} p^{-s} + \frac{1}{2} \sum_{p} p^{-2s} + \frac{1}{3} \sum_{p} p^{-3s} + ...$ (2)

... so that is...

$\ln \zeta(s) - \sum_{p} p^{-s} = \frac{1}{2} \sum_{p} p^{-2s} + \frac{1}{3} \sum_{p} p^{-3s} + ...$ (3)

Now we can proceed expanding in series the ratio $\frac{\zeta(2s)}{1-2^{-s}}$ ... and that is left to You...

Kind regards

$\chi$ $\sigma$
Is...

$\frac{1}{2}\ \frac{\zeta(2s)}{1-2^{-s}} = \frac{1}{2} \sum_{n} n^{-2s} + 2^{-(1+s)}\ \sum_{n} n^{-2s} + 2^{-(1+2s)}\ \sum_{n} n^{-2s} +... >$

$> \frac{1}{2} \sum_{n} n^{-2s} + \frac{1}{2}\ \sum_{n} n^{-3s} + ...$ (1)

... and if you compare the series (1) with the series...

$\ln \zeta(s) - \sum_{p} p^{-s} = \frac{1}{2} \sum_{p} p^{-2s} + \frac{1}{3} \sum_{p} p^{-3s} + ...$ (2)

... it is clear which is greater...

Kind regards

$\chi$ $\sigma$