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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!

wjmccann
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    Your product can also be written as $\prod_j\left(1+\frac1{p_j}\right)$; the infinite product is known to diverge, and good asymptotics for the finite product can be gotten by looking at $\sum_j\log\left(1+\frac1{p_j}\right)$, as you note, and then noting that the value of the log is $\frac1{p_j}+O(p_j^{-2})$, with the sum of the latter terms converging by comparison with the Basel series. – Steven Stadnicki Jan 16 '20 at 21:57
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    This reminds me of perhaps my favorite series to product ever:$$\sum\limits_{n=1}^\infty\frac1n=\prod\limits_{n=1}^\infty\frac{p_n}{p_n-1}$$ – Rushabh Mehta Jan 16 '20 at 22:02
  • To expand on the last part of Steven Stadnicki's comment, this should grow like $C \ln n \ln \ln n$ for some explicit constant $C$. The sum of the reciprocals of primes is controlled by Mertens' theorem. – Erick Wong Jan 16 '20 at 22:36
  • so all possible products of the first n primes +1 divided by the primorial ... –  Jan 17 '20 at 01:11

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[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. $$\Omega(n)=\sum_{p^\alpha |n}\alpha.$$

clathratus
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  • Is there a name for this finite prime zeta function that I could look up, just so I could look up any papers on its asymptotic? – wjmccann Jan 16 '20 at 23:18
  • @wjmccann Any time you have an infinite sum or product, its definition is the limit of finite partial sums or products. It is rare for such parial sums or products to be singled out specifically. One notable example is in this MSE question 136996 "Partial sums of exponential series". – Somos Jan 17 '20 at 00:31
  • @wjmccann As was mentioned in my answer, $\lim_{M\to\infty}\zeta_M(s)=\zeta(s)=\sum_{k\ge1}k^{-s}$, which is the Riemann zeta function. I do not know of a name for $f_M(s)$, but we already know it has a closed form in terms of $\zeta_M$. I am currently in the process of adding to my answer to include more information about these functions, and I will tell you when I have done so. – clathratus Jan 17 '20 at 01:15
  • Okay I have included a partial edit. Expect a more detailed edit in the future. – clathratus Jan 17 '20 at 01:38