TITLE: Evaluating $\sum\limits_{n=1}^{\infty} \frac{1}{n\operatorname{ GPF}(n)}$, where $\operatorname{ GPF}(n)$ is the greatest prime factor
QUESTION [24 upvotes]: $\operatorname{ GPF}(n)=$Greatest prime factor of $n$, eg. $\operatorname{ GPF}(17)=17$, $\operatorname{ GPF}(18)=3$.
How to evaluate convergence/divergence/value of the sum
$$\sum_{n=1}^{\infty} \frac{1}{n\operatorname{ GPF}(n)}\,?$$

REPLY [2 votes]: 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...