Evaluating $\sum\limits_{n=1}^{\infty} \frac{1}{n\operatorname{ GPF}(n)}$, where $\operatorname{ GPF}(n)$ is the greatest prime factor $\operatorname{ GPF}(n)=$Greatest prime factor of $n$, eg. $\operatorname{ GPF}(17)=17$, $\operatorname{ GPF}(18)=3$.
How to evaluate convergence/divergence/value of the sum
$$\sum_{n=1}^{\infty} \frac{1}{n\operatorname{ GPF}(n)}\,?$$
 A: Let $p_k$ be the $k$-th prime number.
$$
\sum_{n\ge1}\frac{1}{n\operatorname{GPF}(n)}=\sum_{k\ge1}\frac{1}{p_k}\sum_{\operatorname{GPF}(n)=p_k}\frac{1}{n}.
$$
If $\operatorname{GPF}(n)=p_k$, then $n=p_1^{i_1}\dots p_{k-1}^{i_{k-1}}\,p_k^{i_k}$ with $i_j\ge0$, $1\le j<k$ and $i_k\ge1$. It folows that
$$
\begin{align*}
\sum_{\operatorname{GPF}(n)=p_k}\frac{1}{n}&=\Bigl(\sum_{i_1\ge0}p_1^{-i_1}\Bigr)\dots\Bigl(\sum_{i_{k-1}\ge0}p_{k-1}^{-i_{k-1}}\Bigr)\Bigl(\sum_{i_k\ge1}p_k^{-i_k}\Bigr)\\
&=\frac{1}{p_k}\,\prod_{i=1}^k\Bigl(1-\frac{1}{p_i}\Bigr)^{-1}.
\end{align*}
$$
From Merten's theorem
$$
\prod_{i=1}^k\Bigl(1-\frac{1}{p_i}\Bigr)^{-1}\sim e^\gamma\,\log p_k\ ,
$$
and the original series has te same character as
$$
\sum_{k\ge1}\frac{\log p_k}{p_k^2} ,
$$
which is convergent.
A: I've attempted to put a numerical lower bound on the sum. Let $S=\sum_{j=1}^\infty 1/jGPF(j)$, where I assume $GPF(1)=1$. Let $r_i$ be the $i$th prime. Then using the geometric series, we have
$S=1+\displaystyle\sum_{\{e_i\}} r_n^{-2}\left[\frac{1}{1-r_n^{-1}}\right]\prod r_i^{-e_i}$
where the sum is over sets of exponents $\{e_i\}$ such that the highest $i$ occurring is $k$, where $0 \le k < n$. The idea here is that most big contributions to $S$ come from terms with fairly small $k$. Let $m=\sum e_i$. Then
$S \ge 1+\displaystyle\sum_{n=1}^\infty r_n^{-2}\left[\frac{1}{1-r_n^{-1}}\right] \left[1+\sum_{m=1}^\infty \sum_{k=1}^{n-1} r_k^{-m} \sum_{\{e_i\}} 1\right]$ ,
where the 1 term before the sum over $m$ is to account for the $m=0$, $k=0$ term. Now let $B(m,k)$ be the number of partitions of $m$ into $k$ nonnegative terms, with order counted as significant, and with the final term being nonzero. Then $B(m,k)={{m+k-1} \choose k}\ge (1+(m-1)/k)^k$, so
$S \ge 1+\displaystyle\sum_{n=1}^\infty r_n^{-2}\left[\frac{1}{1-r_n^{-1}}\right] \left[1+\sum_{m=1}^\infty \sum_{k=1}^{n-1} r_k^{-m}\left(1+\frac{m-1}{k}\right)^k\right]$
The following code, written in the open-source programming language Yacas, is intended to evaluate this expression:
n_max := 100;
m_max := 100;
prec := 8; /* digits of precision */
n := 1;
rn := 1;
s := N(1,prec); /* take GPF(1)=1, so first term=1 */
While (n<=n_max) [
  u := N(1,prec); /* 1=contribution from m=0, k=0 */
  rn := NextPrime(rn); /* rn=nth prime */
  m := 1;
  While (m<=m_max) [
    k := 1;
    rk := 1;
    While (k<n) [
      rk := NextPrime(rk); /* rk=kth prime */
      u := N(u+(1+(m-1)/k)^k*rk^-m,prec);
      k := k+1;
    ];
    m := m+1;
  ];
  sn := N(u*rn^-2*(1/(1-1/rn)),prec);
  s := N(s+sn,prec);
  Write(n,sn,s); NewLine();
  n := n+1;
];
Write(s); NewLine();

Summing up to maximum $n$ and $m$ values of 100 gives $S\ge 1.39$, which is in agreement with, but poorer than, the estimates by J.M. and bgins. 
I've caught and corrected several errors in this after I initially posted it. Maybe there are more...
A: If $\{r_1,\dots,r_k,p\}$ is the set of primes $\le p$, then
$$\{n\in \mathbb{Z}: \operatorname{gpf}(n)=p\}=\{r_1^{e_1}\cdots r_k^{e_k}p^f:\text{each }e_i\ge 0,f\ge1\}.$$
Now we have the Euler product factorization
$$\sum_{e_1\,\ge0}\cdots\sum_{e_k\,\ge0}\sum_{f\ge1}\frac{1}{r_1^{e_1}\cdots r_k^{e_k} p^f}=  \prod_{i=1}^k \left(1+\frac{1}{r_i}+\frac{1}{r_i^2}+\cdots\right)\cdot \left(\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+\cdots\right) $$
$$\implies \bullet =\sum_p \frac{1}{p}\sum_{\operatorname{gpf}(n)=p}\frac{1}{n}=\sum_p \frac{1}{p}\left[\,\prod_{q\,< \,p}\left(1-\frac{1}{q}\right)^{-1}\right]\frac{1/p}{1-1/p}$$
$$=\sum_p \frac{1}{p(p-1)}\prod_{q<p}\left(1-\frac{1}{q}\right)^{-1}.$$
With Mertens' Theorem, we know that the $\prod$ above is $\le C\log p$ eventually, so after we ignore a finite sum corresponding to the $p$ before this "eventually" we may say that
$$\frac{\bullet}{C} + \zeta_P'(2)\le \sum_p \frac{\log p}{p}\left(\frac{1}{p-1}-\frac{1}{p}\right)\le \sum_p\frac{\log p}{p^2}=-\zeta_P'(2).$$
Here $\zeta_P$ is the prime zeta function. With monotonicity of the terms, this proves convergence.
