Infinite product involving primes

Question

I just had my first analysis course as an undergraduate, and I'm trying to learn more about analytic number theory. Right now I'm looking at prime numbers in particular--I'm studying (mostly just thinking and scribbling) on my own during my winter break. I've come across an infinite product that I'm trying to evaluate: I'm fairly sure it converges to 0, but I have no idea of how to prove this, since it is a product over the primes and, so far as I know, we have very few techniques for analyzing the behavior of the prime numbers.

I haven't learned how to use LaTex yet, but the infinite product is simple: it is the product over all primes of (p-1)/p.

It seems intuitive to me that the limit of the partial products is 0, but I have no idea how to prove it, mostly because I don't know how to analyze the behavior of the primes in this setting.

Any help would be very much appreciated. Thanks.

Answer 1

Here is a modified version of proof which I saw in Alan Bakers Book Comprehensive course in Number Theory. In this I have used the prime number theorem but this can be done without it as well as done in the book I mentioned above.

The Prime Number Theorem states that $$\pi (x) = \frac{x}{\log x}+O(\frac{x}{(\log x)^2})$$

First notice that evaluating $\sum\limits_{p \leq x} \ln(1-\frac{1}{p})$ is enough since one has to just take its exponential to obtain the required estimate i.e $\displaystyle\prod \limits_{p \leq x} (\frac{p-1}{p})=\frac{ce^A}{\ln x}$

Mertens Result

$\displaystyle \sum\limits_{p \leq x} \ln(1-\frac{1}{p})= A+\ln(\frac{1}{\ln x})+\ln(c+O(\frac{1}{\ln x}))$

Proof:

We observe that $\displaystyle \sum\limits_{p \leq x} \ln(1-\frac{1}{p})=\sum \limits_{p \leq x} \sum \limits_{m=1}^{\infty} -\frac{1}{mp^m}$ $$\displaystyle \sum \limits_{p \leq x} \sum \limits_{m=1}^{\infty} -\frac{1}{mp^m}=\sum \limits_{p \leq x} \sum \limits_{m=1}^{1} -\frac{1}{mp^m}+\sum \limits_{p \leq x} \sum \limits_{m=2}^{\infty} -\frac{1}{mp^m}$$

But $\displaystyle \sum \limits_{m=2}^{\infty} \frac{1}{mp^m}=O(\frac{1}{p^2})$ this implies that

$$\displaystyle \sum \limits_{p \leq x} \sum \limits_{m=1}^{\infty} -\frac{1}{mp^m}=-\sum \limits_{p \leq x} \frac{1}{p}+\sum \limits_{p} (\ln(1-\frac{1}{p})+\frac{1}{p})-\sum \limits_{p>x} (\ln(1-\frac{1}{p})+\frac{1}{p})$$

$$\displaystyle \sum \limits_{p \leq x} \sum \limits_{m=1}^{\infty} -\frac{1}{mp^m}=-\sum \limits_{p \leq x} \frac{1}{p}+c+O(\sum \limits_{p>x} \frac{1}{p^2})$$ $$\hspace{24mm}=-\sum \limits_{p \leq x} \frac{1}{p}+c+O(\sum \limits_{p>x} \frac{1}{p^2})$$ $$\hspace{18mm}=-\sum \limits_{p \leq x} \frac{1}{p}+c_1+O(\frac{1}{x})$$

Now we will try to estimate the sum $-\sum \frac{1}{p}$

Here I have proved it asssuming the Prime Number Theorem

$\displaystyle \sum\limits_{n \leq x} a_n f(n)=s(x)f(x) - \int\limits_{1}^{x} s(u)f'(u)du$. Now taking $a_n=1$ if $n$ is a prime and $0$ otherwise and taking $f(x)=\frac{1}{x}$ we obtain $$\displaystyle\sum \limits_{p \leq x} \frac{1}{p}= \frac{\pi(x)}{x}+\int \limits_{1}^{x}\frac{\pi(u)}{u^2} du$$ From this after integrating we obtain $$\displaystyle\sum\limits_{p} \frac{1}{p}=\ln \ln x +c_2 + O(\frac{1}{\ln x})$$ Hence Proved.

Answer 2

A more elementary way to prove this would be to notice that one may write $$\left(1+\frac{1}p+\frac{1}{p^2}+\frac{1}{p^3}+\ldots\right)=\frac{1}{1-\frac{1}p}=\frac{p}{p-1}.$$ This is the reciprocal of the term you want - which is to say that the product $$S=\prod_p 1+\frac{1}p+\frac{1}{p^2}+\frac{1}{p^3}+\ldots$$ is the reciprocal of the one you're finding (since $\frac{1}x\frac{1}y=\frac{1}{xy}$ we can bring the reciprocal outside the product). We want to show that $S=\infty$. However, this isn't too hard: We can use use a sort of distributive law to turn the above into a sum - in particular, notice that, for instance $$\left(1+\frac{1}2+\frac{1}{2^2}+\ldots\right)\left(1+\frac{1}3+\frac{1}{3^2}+\ldots\right)$$ will be the sum $$1+\frac{1}{2}+\frac{1}{2^2}+\ldots +\frac{1}3+\frac{1}{3\cdot 2}+\frac{1}{3\cdot 2^2}+\ldots + \frac{1}{3^2}+\frac{1}{3^2\cdot 2}+\frac{1}{3^2\cdot 2^2}\ldots$$ where the sum runs over all numbers of the form $\frac{1}{2^a3^b}$. Extending to the infinite case, we can show that, when we distribute out the product for $S$, we get the sum of $\frac{1}n$ over all integers writable as a product of primes - since all positive integers are uniquely representable as such a product, we can conclude that $$S=\sum_{n=1}^{\infty}\frac{1}n$$ meaning $S$ diverges to $\infty$ so $\frac{1}S$ is $0$.

You can make this rigorous by noting the following equality relating partial sums: $$\sum_{n=1}^{k}\frac{1}n\leq \prod_{p\leq k}\left(1+\frac{1}p+\frac{1}{p^2}+\ldots + \frac{1}{p^k}\right)$$ which follows from your run-of-the-mill distributive law and the fact that all $n$ below $k$ are writable as a product of primes below $k$ with exponent less than $k$.