A semiprime $s$ is a positive integer that is the product of two prime numbers, see Semiprine from the encyclopedia Wikipedia, thus corresponding to the sequence A001358 of the OEIS. I wondered if it is possible to deduce something about the veracity of the following conjecture.
Conjecture. There exist a positive integer $n_0$ and a positive constant $C$ such that the following inequality holds $$\Bigl(\prod_{\substack{1\leq s\leq n\\s\text{ semiprime}}}s\Bigr)\Bigl(\sum_{\substack{1\leq s\leq n\\s\text{ semiprime}}}\frac{1}{s}\Bigl)\leq C e^n\tag{1}$$ for all integers $n>n_0$.
On my computation (with my computer), if there are no mistakes I've that $\frac{1}{n}\log\Bigl(\Bigl(\prod_{\substack{1\leq s\leq n\\s\text{ semiprime}}}s\Bigr)\Bigl(\sum_{\substack{1\leq s\leq n\\s\text{ semiprime}}}\frac{1}{s}\Bigl)\Bigl)\leq c$, for the segment of intergers $6\leq n\leq 6000$ where the constant $c$ seems to tend to $\approx 2$ for previous segment of integers. These computations were the motivation for our Conjecture.
Question. I would like to know what work can be done with the purpose to prove or refute previous conjecture. If the inequality makes sense you can to improve it to get a more sharp expression (as an inequality or an asymptotic identity). Many thanks.
I don't know if this conjecture is in the literature, to ask this question I was inspired in a statement from [1]. If in the literature are expressions for the LHS of $(1)$ that answer explicitely my question, then refer it answering my question as a reference request and I try to search an read those statements from the literature.
References:
[1] Takashi Agoh, Paul Erdös and Andrew Granville, Primes at a (Somewhat Lengthy) Glance, The American Mathematical Monthly, Vol. 104, No. 10 (December, 1997), pp. 943-945.
The conjecture does not hold, the left hand side of $(1)$ grows faster than $e^{a\cdot n}$ for every $a \in \mathbb{R}$. But one may need to use large $n$ to actually see that effect.
Any Chebyshev-type bound $$\vartheta(x) \geqslant b\cdot x \tag{1}$$ for all $x \geqslant x_b$, where $b > 0$ and $\vartheta(x)$ is the sum of the logarithms of the primes not exceeding $x$, together with the divergence of the series of reciprocals of the primes, suffices to see that.
Given a $b$ with $(1)$ and an arbitrary $a \in \mathbb{R}$, choose $y$ such that $$b\cdot \sum_{p \leqslant y} \frac{1}{p} > a + 1\,. \tag{2}$$ Then let $n > y\cdot (y + x_b)$. Among the semiprimes $\leqslant n$, consider only those whose smallest prime factor is $\leqslant y$. For every prime $p \leqslant y$, the product of the semiprimes $\leqslant n$ whose smallest prime factor is $p$ is \begin{align} p^{\pi(n/p) - \pi(p-1)}\cdot \exp\bigl(\vartheta(n/p) - \vartheta(p-1)\bigr) &\geqslant \exp\bigl(\vartheta(n/p) - \vartheta(y)\bigr) \\ &\geqslant \exp \biggl(b\cdot \frac{n}{p} - \vartheta(y)\biggr)\,, \end{align} hence the product of all semiprimes $\leqslant n$ whose smallest prime factor is $\leqslant y$ is at least $$\exp \Biggl( n\cdot b\sum_{p \leqslant y} \frac{1}{p} - \pi(y)\vartheta(y)\Biggr) \geqslant e^{a\cdot n}\cdot \exp\bigl( n - \pi(y)\vartheta(y)\bigr)\,.$$ If $n$ is also not smaller than $\pi(y)\vartheta(y)$ [which is automatically the case since we required $n \geqslant y^2$, and that is $> \pi(y)\vartheta(y)$ for all $y > 0$], this product is therefore not smaller than $e^{an}$. The neglected semiprimes whose smallest prime factor is larger than $y$ make the overall product of semiprimes larger, and the sum of the reciprocals of the semiprimes $\leqslant n$ is $> 1$ for sufficiently large $n$, thus the left hand side of $(1)$ is larger than $e^{an}$ for all large enough $n$, regardless of the value of $a$.