# Convergence

• March 30th 2011, 09:42 AM
usagi_killer
Convergence
Does $\displaystyle{\prod_{\text{primes p}}\left(1+\frac{1}{p}\right)}$ converge, and if so to what value?

• March 30th 2011, 10:25 AM
chisigma
The Swiss mathematician Leonhard Euler demonstrated two and half centuries ago that the series...

$\displaystyle \sum_{p} \frac{1}{p}$

... diverges, so that diverges also the product...

$\displaystyle \prod_{p} (1+\frac{1}{p})$

Kind regards

$\chi$ $\sigma$
• March 30th 2011, 10:50 AM
FernandoRevilla
However, if $s>1$ then,

$\displaystyle\prod_{p}\left(1+\dfrac{1}{p^s}\right )=\dfrac{\zeta(s)}{\zeta{(2s)}}$

So, can extend the formula,

$\displaystyle\prod_{p}\left(1+\dfrac{1}{p}\right)= \dfrac{\zeta(1)}{\zeta{(2)}}=\dfrac{+\infty}{\pi^2/6}=+\infty$
• March 30th 2011, 09:08 PM
usagi_killer
@chisigma thanks for the help! however I am still a bit unsure on how the divergence of the sum implies that the product also diverges?

@FernandoRevilla cheers for that! I haven't seen that formula yet, nor can I find it in my notes, however im very interested, can you perhaps show me a proof of how the formula comes about?

thanks again!
• March 30th 2011, 09:41 PM
FernandoRevilla
Quote:

Originally Posted by usagi_killer
@FernandoRevilla cheers for that! I haven't seen that formula yet, nor can I find it in my notes, however im very interested, can you perhaps show me a proof of how the formula comes about?

We have for every $s>1$:

$\dfrac{\zeta (s)}{\zeta (2s)}=\dfrac{\prod_{p} (1-p^{-s})^{-1}}{\prod_{p} (1-p^{-2s})^{-1}}=\ldots$
• March 31st 2011, 02:13 AM
chisigma
Quote:

Originally Posted by usagi_killer
@chisigma thanks for the help! however I am still a bit unsure on how the divergence of the sum implies that the product also diverges?

In...

Convergence of Infinite Products « The Everything Seminar

... there is the proof of the following theorem: an 'infinite product' of the form...

$\displaystyle \prod_{n=1}^{\infty} (1+a_{n})$

... converges if and only if the 'infinite sum'...

$\displaystyle \sum_{n=1}^{\infty} a_{n}$

... converges
...

Kind regards

$\chi$ $\sigma$
• March 31st 2011, 02:57 AM
usagi_killer
Thanks guys! Makes sense now!