On the convergence of a series$ \sum_{n=1}^{\infty}\frac{1}{n W(n)}$

Question

How to show convergence of $$ \sum_{n=1}^{\infty}\frac{1}{n W(n)}$$ where $W(n)$ denotes the product of distinct primes dividing n.

Answer 1

If we consider some finite subset $E$ of the prime numbers and restrict the given series to the natural numbers divided by each prime in $E$ (let we call this subset of $\mathbb{N}$ as $E'$), we get

$$ \sum_{n\in E'}\frac{1}{n W(n)} = \frac{1}{\prod_{p\in E}p}\sum_{n\in E'}\frac{1}{n}=\frac{1}{\left(\prod_{p\in E}p\right)^2}\prod_{p\in E}\frac{1}{1-\frac{1}{p}}=\prod_{p\in E}\frac{1}{p(p-1)} $$ hence the sum restricted to the integers that are divided by some prime in $E$ equals $\prod_{p\in E}\left(1+\frac{1}{p(p-1)}\right)$, and by taking the limit on both sides as "$E$ tends to the set of all primes", which is valid because everything is nonnegative: $$ \sum_{n\geq 1}\frac{1}{n W(n)}=\prod_{p}\left(1+\frac{1}{p(p-1)}\right) $$ and the last infinite product is convergent since $\sum_{p}\frac{1}{p(p-1)}$ is convergent, but bounded below by $$ \prod_{p}\left(1-\frac{1}{p^2}\right)^{-1}=\zeta(2).$$ On the other hand, we may also compute it exactly: $$ \prod_{p}\left(1+\frac{1}{p(p-1)}\right)=\prod_{p}\left(1-\frac{1}{p^2}\right)^{-1}\left(1+\frac{1}{p^3}\right)=\color{red}{\frac{\zeta(2)\,\zeta(3)}{\zeta(6)}}=1.94359643682\ldots $$