# Prime products

eljose
Is this equaltiy exact?:

$$\Pi(a_p+b_p)= \Pi(a_p)+\Pi(b_p)$$

where both products a_p,b_p and (a_p+b_p) converge

another qeustion $$\Pi 1=1$$ ?

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Homework Helper
your first conjecture is vacuous i think, though i haven't checked properly: there is no situation where all three products can exist for the usual definitions of convergent products, namely, prod(1+x_n) converges iff sum(x_n) converges. and besides, does it even seem reasonable? if the index set were finite it is false.

and yes for the second one.

Staff Emeritus
Gold Member
The equality is not exact -- it's not in the ballpark. It's impossible for all three of those infinite products to converge.

Yes, the infinite product of the constant 1 is, in fact, 1, as can be easily seen by taking the limit of the partial products.

eljose
how about $$a_p=1 , b_p=\frac{1}{1-p^{-s}}$$they both converge, if the equality is exact the sum:

$$a_p-b_p=\frac{p^{-s}}{1-p^{-s}}$$ the product would converge to

$$1-\zeta(s)$$ being R(s) riemann,s zeta function..is that true?..

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Homework Helper
did you not read the posts above?

eljose
yes i have read the post, i have given two converging products and i would like to know if the equality $$1-\zeta(s)=\Pi\frac{p^{-s}}{1-p^{-s}}$$ is exact, because you can see that all the products

a(p)=1 b(p)=1/1-p^{-s} and a(p)+b(p) converge for certain values of s and how about?...

$$1+1/\zeta(s)=\Pi(2+p^{-s}$$

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Homework Helper
so, you'vwe been told that if prod(1+x_n) is convergent, and so is prod(1+y_n) then the prod(1+x_n+1+y_n) cannot be convergent. so why are you still asking if it is possible?

let p+n be the n'th prime.

let 1+y_n = a_{p_n}-b_{p_n} and 1+x_n= b_{p_n}. you are claiming these both converge, therefore, for whatever value of s this correpsonds to, to a_{p_n} cannot converge. but a_p_n=1 so actually (at least one of) the other two products cannot exist.

eljose
could you help me with these product and say if they converge and what would be their value?..

$$\Pi\frac{2+p^{-s}}{1+p^{-s}} \Pi2-p^{-s}$$

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Homework Helper
it converges iff the sum of 1/(1+p^{-s}) work out when this can happen, eg by the sandwich principle say. i think you'll find it's approximately when s<-1 (assuming s real) which as we know is out of the range of the region of convergence for the zeta function.

edit: now you've changed it. and $\prod 2$ doesn't make sense, really. what's the insdex? why is there a stray p^-s involved? why am i bothering to help you since you never listen to me anyway?

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Homework Helper
matt grime said:
it converges iff the sum of 1/(1+p^{-s}) work out when this can happen, eg by the sandwich principle say. i think you'll find it's approximately when s<-1 (assuming s real) which as we know is out of the range of the region of convergence for the zeta function.

edit: now you've changed it. and $\prod 2$ doesn't make sense, really. what's the insdex? why is there a stray p^-s involved? why am i bothering to help you since you never listen to me anyway?
Reading his code, not that it really matters anyway, it's meant to be displayed as:

$$\prod \left(\frac{2+p^{-s}}{1+p^{-s}}\right) \prod \left(2 - p^{-s}\right)$$

Although to me it still seems quite ill-defined, but never mind.

$$1+1/\zeta(s)=\prod f(p)$$