We already have a definition plus two expressions for a real number $\rho$, including a $\zeta$-formula provided by @Lucia:

$$\rho\ :=\ \sum_{n\in\mathbb N}\,\frac 1{n\cdot rad(n)}\ = \ \prod_{p\in\mathbb P}\,\left(1+\frac 1{p\cdot (p-1)}\right) \ =\ \frac {\zeta(2)\cdot\zeta(3)}{\zeta(6)} $$

Let $\rho$ enjoy another on. In addition to function $rad(n)$ we will need an auxiliary function $rad'(n)$:

$$ rad(n)\ :=\ \prod\{p\in\mathbb P: p|n\}\\ rad'(n)\ := \prod\{p-1: p\in rad(n)\} $$

for all $\ n\in\mathbb N.\ $ Next, let:

$$ \mathbf {Rad}\ :=\ \{n\in\mathbb N: rad(n) = n\} $$

THEOREM

$$ \rho\ =\ \sum_{r\,\in\,\mathbf Rad}\, \frac 1{r\cdot r'} $$

where $\ r' = rad'(r)$.

REMARK $\ rad'(1) = 1\ $ due to the Bourbaki kind of a convention.

PROOF

$$\rho\ :=\ \sum_{n\in\mathbb N}\,\frac 1{n\cdot rad(n)}\ =\ \sum_{r\in\mathbf{Rad}}\,\left(\frac 1{r^2}\cdot\prod_{p|r}\frac p{p-1}\right) $$ $$ =\ \sum_{r\in\mathbf {Rad}}\,\frac 1{r\cdot r'} $$

END of proof

We already have a definition plus two expressions for a real number $\rho$, including a $\zeta$-formula provided by @Lucia:

$$\rho\ :=\ \sum_{n\in\mathbb N}\,\frac 1{n\cdot rad(n)}\ = \ \prod_{p\in\mathbb P}\,\left(1+\frac 1{p\cdot (p-1)}\right) \ =\ \frac {\zeta(2)\cdot\zeta(3)}{\zeta(6)} $$

Let $\rho$ enjoy another on. In addition to function $rad(n)$ we will need an auxiliary function $rad'(n)$:

$$ rad(n)\ :=\ \prod\{p\in\mathbb P: p|n\}\\ rad'(n)\ := \prod\{p-1: p\in rad(n)\} $$

for all $\ n\in\mathbb N.\ $ Next, let:

$$ \mathbf {Rad}\ :=\ \{n\in\mathbb N: rad(n) = n\} $$

THEOREM

$$ \rho\ =\ \sum_{r\,\in\,\mathbf Rad}\, \frac 1{r\cdot r'} $$

where $\ r' = rad'(r)$.

REMARK 1 $\ rad'(1) = 1\ $ due to the Bourbaki kind of a convention.

PROOF

$$\rho\ :=\ \sum_{n\in\mathbb N}\,\frac 1{n\cdot rad(n)}\ =\ \sum_{r\in\mathbf{Rad}}\,\left(\frac 1{r^2}\cdot\prod_{p|r}\frac p{p-1}\right) $$ $$ =\ \sum_{r\in\mathbf {Rad}}\,\frac 1{r\cdot r'} $$

END of proof

REMARK 2 Hm, equality

$$ \prod_{p\in\mathbb P}\,\left(1+\frac 1{p\cdot (p-1)}\right)\ =\ \sum_{r\,\in\,\mathbf Rad}\, \frac 1{r\cdot r'} $$

is immediate.

We already have a definition plus two expressions for a real number $\rho$, including a $\zeta$-formula provided by @Lucia:

$$\rho\ :=\ \sum_{n\in\mathbb N}\,\frac 1{n\cdot rad(n)}\ = \ \prod_{p\in\mathbb P}\,\left(1+\frac 1{p\cdot (p-1)}\right) \ =\ \frac {\zeta(2)\cdot\zeta(3)}{\zeta(6)} $$

Let $\rho$ enjoy another on. In addition to function $rad(n)$ we will need an auxiliary function $rad'(n)$:

$$ rad(n)\ :=\ \prod\{p\in\mathbb P: p|n\}\\ rad'(n)\ := \prod\{p-1: p\in rad(n)\} $$

for all $\ n\in\mathbb N.\ $ Next, let:

$$ \mathbf {Rad}\ :=\ \{n\in\mathbb N: rad(n) = n\} $$

THEOREM

$$ \rho\ =\ \sum_{r\,\in\,\mathbf Rad}\, \frac 1{r'} $$

where $\ r' = rad'(r)$.

REMARK $\ rad'(1) = 1\ $ due to the Bourbaki kind of a convention.

PROOF

$$\rho\ :=\ \sum_{n\in\mathbb N}\,\frac 1{n\cdot rad(n)}\ =\ \sum_{r\in\mathbf{Rad}}\,\left(\frac 1r\cdot\prod_{p|r}\frac p{p-1}\right) $$ $$ =\ \sum_{r\in\mathbf {Rad}}\,\frac 1{r'} $$

END of proof

We already have a definition plus two expressions for a real number $\rho$, including a $\zeta$-formula provided by @Lucia:

$$\rho\ :=\ \sum_{n\in\mathbb N}\,\frac 1{n\cdot rad(n)}\ = \ \prod_{p\in\mathbb P}\,\left(1+\frac 1{p\cdot (p-1)}\right) \ =\ \frac {\zeta(2)\cdot\zeta(3)}{\zeta(6)} $$

Let $\rho$ enjoy another on. In addition to function $rad(n)$ we will need an auxiliary function $rad'(n)$:

$$ rad(n)\ :=\ \prod\{p\in\mathbb P: p|n\}\\ rad'(n)\ := \prod\{p-1: p\in rad(n)\} $$

for all $\ n\in\mathbb N.\ $ Next, let:

$$ \mathbf {Rad}\ :=\ \{n\in\mathbb N: rad(n) = n\} $$

THEOREM

$$ \rho\ =\ \sum_{r\,\in\,\mathbf Rad}\, \frac 1{r\cdot r'} $$

where $\ r' = rad'(r)$.

REMARK $\ rad'(1) = 1\ $ due to the Bourbaki kind of a convention.

PROOF

$$\rho\ :=\ \sum_{n\in\mathbb N}\,\frac 1{n\cdot rad(n)}\ =\ \sum_{r\in\mathbf{Rad}}\,\left(\frac 1{r^2}\cdot\prod_{p|r}\frac p{p-1}\right) $$ $$ =\ \sum_{r\in\mathbf {Rad}}\,\frac 1{r\cdot r'} $$

END of proof