Estimating product of the primes dividing N.

Question

I am running through a proof by Stefan A. Burr which uses the result that is as follows. Where $p$ is a prime and $\text{N}\in\mathbb{Z}$:

$$\prod_{p|N}{(1+1/p)} = O(\text{log(log(3N)})$$

Can anyone explain to me why this is the case?

Answer 1

First, note that we can assume that $N$ is the product of all the primes up to some parameter $x$: Let $k$ be the number of distinct primes dividing $N$, and let $N'$ be the product of the first $k$ primes. Then $\prod_{p\mid N} (1+\frac1p) \le \prod_{p\mid N'} (1+\frac1p)$, since each product is over $k$ primes, and the $j$th prime occurring on the right-hand side is at most as large as the $j$th prime occurring on the left-hand side (and the function $p\mapsto1+\frac1p$ is decreasing, and repeated prime factors on the left-hand side are irrelevant by definition). A similar argument shows that $\log\log(3N') \le \log\log(3N)$. Therefore proving $\prod_{p\mid N'} (1+\frac1p) = O(\log\log(3N'))$ implies that $\prod_{p\mid N} (1+\frac1p) = O(\log\log(3N))$.

So we have reduced to proving $$ \prod_{p\le x} \bigg( 1+\frac1p \bigg) = O\bigg( \log\log\bigg( 3\prod_{p\le x} p \bigg) \bigg). \tag{*} $$ By Mertens's theorems (and the value of $\zeta(2)$), the left-hand side is asymptotic to $\frac{6e^\gamma}{\pi^2} \log x$. The product on the right-hand side is exactly $e^\theta(x)$, and $\theta(x)\sim x$ by the prime number theorem. Therefore the right-hand side is asymptotic to $\log x$. These two asymptotic evaluations establish (*) for large $x$, but of course we can adjust the $O$-constant to confirm it for all $x$.

Answer 2

By Mertens' second theorem $$ \sum_{p\leq n}\frac{1}{p}\sim \log\log(n)\tag{1} $$ hence when dealing with $\sum_{p\mid N}\frac{1}{p}$ we may just consider the primes dividing $N$ and being $\leq \log N$, then the primes dividing $N$ and being $\geq \log N$. The contribute given by large primes to $\sum_{p\mid N}\frac{1}{p}$ is bounded by a constant and the contribute given by small primes is bounded by $\log\log\log(N)$ by $(1)$. It follows that $$ \prod_{p\mid N}\left(1+\frac{1}{p}\right)=O\left(\exp\sum_{p\mid N}\frac{1}{p}\right)=O(\log\log N).\tag{2} $$