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.
Best Answer
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 $k$.