Does $\displaystyle \displaystyle{\prod_{\text{primes p}}\left(1+\frac{1}{p}\right)}$ converge, and if so to what value?

I'm not too sure how to go about this question, any help would be greatly appreciated!

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

I'm not too sure how to go about this question, any help would be greatly appreciated! - Mar 30th 2011, 10:25 AMchisigma
The Swiss mathematician Leonhard Euler demonstrated two and half centuries ago that the series...

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Mar 30th 2011, 10:50 AMFernandoRevilla
However, if $\displaystyle s>1$ then,

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

So, can extend the formula,

$\displaystyle \displaystyle\prod_{p}\left(1+\dfrac{1}{p}\right)= \dfrac{\zeta(1)}{\zeta{(2)}}=\dfrac{+\infty}{\pi^2/6}=+\infty$ - Mar 30th 2011, 09:08 PMusagi_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! - Mar 30th 2011, 09:41 PMFernandoRevilla
- Mar 31st 2011, 02:13 AMchisigma
In...

Convergence of Infinite Products « The Everything Seminar

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

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

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

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

... converges

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Mar 31st 2011, 02:57 AMusagi_killer
Thanks guys! Makes sense now!