Let $\ \mathbb N:= \{1\ 2\ \ldots\}\ $ be the set of natural numbers. Let $\ \mathbb P:=\{2\,\ 3\,\ 5\,\ 7\,\ 11\,\ \ldots\}\ $ be the set of primes. Then natural radical $\ rad(n)\ $ is

$$ rad(n)\ :=\ \prod\,\{p\in\mathbb P: p\,|\,n\} $$

for every $\ n\in\mathbb N$. Define

$$ \rho\ :=\ \sum_{N\in\mathbb N}\,\frac 1{n\cdot rad(n)} $$

hence $\ \rho > \frac{\pi^2}6.\ $ Furthermore, it'd be an exercise for 1' year college students (or advanced high school students) to show that $\ \rho<\infty.\ $ Don't tell them too early that

$$\ \rho\ =\ \prod_{p\in \mathbb P}\,\left(1+\frac 1{p\cdot(p-1)}\right) $$

QUESTION $\ $ Can you find (yourself or a reference) an algerbaic expression, possibly in terms of $\pi$ and/or $e$ for $\rho$.

Something like $\ \frac{\pi^{\frac 52}}6$ or similar.

$$ \rho(s)\ :=\ \sum_{n\in\mathbb N}\,\frac 1{n\cdot (rad(n))^s} $$

for every complex $\ s\in\mathbb C\ $ such that it's real part is positive,$\ \Re(s)>0.\ $ Thus

$$ \rho(s)\ =\ \prod_{p\in\mathbb P}\, \left(1+\frac 1{p^s\cdot(p-1)}\right) $$

QUESTION A $\ $What would be the compact expressions for $\ \rho(n)\ $ for all or as many natural values of $\ n\ $ as possible?

QUESTION B $\ $ What are the other formulas for $\ \rho(s)\ $ in the style of multiplicative number theory.

QUESTION C $\ $ How does the complex analytic extension of $\ \rho(s)\ $ look like?

$\qquad\qquad\qquad$ NOTES

Here is my third personal proof (:-) of the infinitude of primes:

THEOREM (Euclid) $\ |\mathbb P| = \infty$

PROOF $\ $ We see from the product representation of $\ \rho\ $ that $\ \rho<\infty\ $ in general, and especially if $|\mathbb P|<\infty.\ $

On the other hand, when $\ |\mathbb P|<\infty$ then $\ r:=\prod\mathbb P\in\mathbb N.\ $ Then

$$ \rho\ >\ \sum_{k=1}^n\,\frac 1{k\cdot rad(k)}\ \ge\ \frac 1r \cdot\sum_{k=1}^n\,\frac 1k\ \longrightarrow\ \infty $$

when $\ n\rightarrow\infty.\ $ Assumption $\ |\mathbb P|<\infty\ $ has lead us to a contradiction.

END of proof

$\qquad\qquad\qquad$ ADDENDUM

Let me copy @Lucia's formula from their first comment below. You may also read Lucia's second comment related to a paper by Bateman. In the derivation below I applied @Wojowu's observation from a comment below, which has drastically simplified and shortened one of the steps.

THEOREM (@Lucia) $$ \rho\ =\ \frac {\zeta(2)\cdot\zeta(3)}{\zeta(6)} $$

PROOF

$$ \frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}\ = \ \frac{p^3+1}{p\cdot(p^2-1)} \cdot \frac{p^6}{p^6-1}\ =\

\frac{p^2-p+1}{p\cdot(p-1}} \cdot \frac{p^6}{p^6-1}\ = \ \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\ $$

i.e. $$ \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\ = \ \frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}$$

Thus $$ \prod_{p\in\mathbb P}\, \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\ = \ \prod_{p\in\mathbb P}\,\frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}$$ or $$ \rho\cdot\zeta(6)\ =\ \zeta(2)\cdot\zeta(3) $$

and the theorem follows. END of proof

END of Proof