Prove the product of a sum of powers of primes diverges

I need to prove that this is divergent: $$\prod_{\stackrel{p\leq N}{p\text{ prime}}}\left(1 + \frac{1}{p} + \frac{1}{p^2} + \cdots + \frac{1}{p^k} + \cdots\right),$$ where the expression inside of the product is: $$\frac{1}{(1-\frac{1}{p})} = (1 + \frac{1}{p} + \frac{1}{p^2} + \cdots + \frac{1}{p^k} + \cdots).$$

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I'm trying to use an integral test, but I'm a little confused because I'm only able to use prime numbers (due to another proof I'm working on), so I can only think of using a Riemann sum, but that's not exactly a proof. –  user7435 Mar 27 '11 at 15:30
That was not a correct edit –  user7435 Mar 27 '11 at 15:32
I am probably wrong, but can someone tell me why the product is not $\zeta(-1)$ –  Please Delete Account Mar 27 '11 at 15:40
@Approximist: You mean $\zeta(1)$, and in that case yes, it does converge to $\zeta(1)$ and hence diverges to infinity. –  Eric Naslund Mar 27 '11 at 16:00
I removed the calculus tag. –  Eric Naslund Mar 27 '11 at 16:02

The fastest way is to use the Fundamental Theorem of Arithmetic (Unique factorization into primes). You can show that in a combinatorial sense $$\prod_p \left( 1+\frac{1}{p}+\frac{1}{p^2}+\cdots\right)=\sum_{n=1}^\infty \frac{1}{n}.$$ I mean that if the product was expanded, each term on the right would appear exactly once. This then proves the divergence of your product since the harmonic series diverges. This is also known as Eulers Proof of the infinitude of the primes.

Hope that helps,

Mertens Estimate: If you are curious, we can also say things about the rate of divergence of this product. It is one of Mertens Estimates which says $$\prod_{p\leq N} \left(1-p^{-1}\right)^{-1} = e^\gamma \ln N + O(1)$$ where $\gamma$ is the Euler-Mascheroni Constant. This means that the product you are looking at grows roughly at the same rate as the logarithm. This is to be expected from the first part since the divergence of the harmonic series is also like the logarithm, and this argument can be made rigorous. However the constant requires slightly more work to compute.

Added: This Wikipedia page is related, and a little interesting too.

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FTA says any integer greater than 1 can be written as a unique product (up to ordering of the factors) of prime numbers How do I know that my product is an integer? –  user7435 Mar 27 '11 at 15:40
@Joseph It's not, it's 1 divided by an integer. Each term will be of the form $1/( p_1^{e_1}p_2^{e_2} ...) = 1/n$ where n is given by its factorisation. (Also note that terms with an 'infinite number of $p_i$'s are zero). –  Myself Mar 27 '11 at 15:45
We don't and in fact it never is an integer. Use the unique factorization on the denominators, and think about expanding the product. (To make sense of expanding a divergent product, what we are doing is assuming it converges, and arriving at a contradiction when the harmonic series appears) The first term in the expansion is a 1, and each term we get will be of the form $\frac{1}{m}$ for some integer $m$. What you have to say is why every integer $m$ will appear once, and only once. Thats where unique factorization comes into play. –  Eric Naslund Mar 27 '11 at 15:51
Every possible product of primes appears exactly once. For example, do you see why $\frac{1}{2\cdot 3^2}$ will appear? What about an arbitrary $$\frac{1}{p_1^{\alpha_1}\cdots p_r^{\alpha_r}}.$$ Why will this appear in the product? Can it appear more then once? –  Eric Naslund Mar 27 '11 at 16:10
Well we have in a sense $$(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots )$$ $$\times (1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots )$$ $$\times (1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\cdots)$$ $$\times\cdots$$ So the term obtained by multiplying $\frac{1}{2}$ and $\frac{1}{3^2}$ and then infinitely many $1$'s will give $\frac{1}{2\cdot 3^2}$. –  Eric Naslund Mar 27 '11 at 16:25

Hint: Expand the partial product using all primes $p\leq N$ for some large $N$, and compare with the partial sum of the harmonic series.

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Hints: 1)inside the parentheses you have a geometric series which can be summed. 2) $\sum_{p\text{ prime}} \frac{1}{p}$ diverges

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I'm not allowed to use this fact in my proof. I have to prove that that sum diverges if I use it. –  user7435 Mar 27 '11 at 15:35
Isn't this just unique factorization plus the fact the harmonic series diverges? –  Uticensis Mar 27 '11 at 15:37
@Ross: And why is (2) true? You have pushed the details into a statement which is more difficult to prove than the original. –  Eric Naslund Mar 27 '11 at 15:39
@Eric: It makes sense that it would be more difficult to prove, but it is "well known". I didn't know what was available to OP, but he says this is not. Your answer is a good one. –  Ross Millikan Mar 27 '11 at 15:50
@Ross: Thanks. Heres an interesting tid-bit about how these results were originally derived. Historically, Euler was the first to prove that $\sum_p \frac{1}{p}$ diverges. (1737) He proved this by showing first that the above product diverges because of its relationship to the harmonic series, and then by showing that any of the higher power terms did not contribute more than a constant. This then implies that the sum of the reciprocals diverges. Also see Wikipedia's page on this (the Erdos proof is cool): en.wikipedia.org/wiki/… –  Eric Naslund Mar 27 '11 at 15:58