I am trying to learn more information about the product series as follows $$ \prod_{j=1}^n\dfrac{p_j + 1}{p_j} $$ where $p_1,p_2,p_3,\ldots,p_n$ are the first $n$ prime numbers. I have no idea where to start looking up information about this.
From what I've seen I now know that the product on the bottom is the primorial function, and it appears to grow very, very slowly, however I am struggling to bound the equation and was wondering if there was prior work done on it.
For example trying to bound it as follows
$$ \prod_{j=1}^n\dfrac{p_j + 1}{p_j} < \log \prod_{j=1}^n p_n $$ $$ \prod_{j=1}^n{(p_j + 1)} < \prod_{j=1}^n p_n \sum_{j=1}^n \log p_n $$
and I hit a wall. If anyone knows any information about this series please let me know!
[converted from a comment]
Let $$\zeta_M(s)=\prod_{p\le M}\frac{1}{1-p^{-s}}$$ where the product runs over all primes $p\le M$. Additionally we define $$f_M(s)=\prod_{p\le M}\frac1{1+p^{-s}}.$$ Then $$\zeta_M(s)f_M(s)=\prod_{p\le M}\frac1{(1+p^{-s})(1-p^{-s})}=\prod_{p\le M}\frac{1}{1-p^{-2s}}=\zeta_M(2s),$$ so that $$f_M(s)=\frac{\zeta_M(2s)}{\zeta_M(s)}.$$ Your product is given by $P_M=1/f_M(1)=\zeta_M(1)/\zeta_M(2)$, which diverges as $M\to\infty$. To see this, note that $\lim_{M\to\infty}\zeta_M(s)=\zeta(s)=\sum_{k\ge1}k^{-s}$ diverges as $s\to 1$, while $\lim_{M\to\infty}\zeta_M(2)=\pi^2/6$.
I leave it to other users to find good asymptotic representations of $P_M$.
Partial edit.
From here we have $$f(s)=\frac{\zeta(2s)}{\zeta(s)}=\sum_{n\ge1}\frac{(-1)^{\Omega(n)}}{n^s},$$ where $\Omega(n)$ is the number of prime factors of $n$ counting multiplicity, i.e. $$\Omega(n)=\sum_{p^\alpha |n}\alpha.$$