Convergence of $\sum_{k=1}^\infty\frac{\operatorname{rad}(k!)}{\operatorname{lcm}(1,2,\ldots,k)}$ I did experiments with a Pari/GP program that suggest that the numerical series 
$$\sum_{k=1}^\infty\frac{\operatorname{rad}(k!)}{\operatorname{lcm}(1,2,\ldots,k)}\tag{1}$$
is convergent, where for an integer $n>1$ $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p \text{ prime}}}p$$
is the product of distinct prime factors dividing $n$, with the definition $\operatorname{rad}(1)=1$. You can see the Wikpedia Radical of an integer to see the properties of such arithmetic function (
it is the famous function of the abc conjecture). We know the size/asymptotic of the least common multiple of the first $n$ integers $\operatorname{lcm}(1,2,\ldots,n)$ and its relationship to the so-called second Chebyshev function, see for example this MathWorld's article Chebyshev Functions.

Question. Provide help, details or hints to prove that our series $$\sum_{k=1}^\infty\frac{\operatorname{rad}(k!)}{\operatorname{lcm}(1,2,\ldots,k)}$$
  is convergent. Many thanks.

Remarks (About that seems that previous Question isn't obvious using comparisons). 
1) Notice that the suare-free kernel $\operatorname{rad}(n)$ is an arithmetic function satisfying that $\operatorname{rad}(n)\leq n$, but the ratio test implies that $\sum_{k=A}^\infty k!e^{-k}$ is divergent, for each fixed positive integer $A$. 
2) We know also Legendre's formula to calculate to evaluate the $p$-adic valuation of $n$, I mean the symbol $\nu_p(n!)$. 
Computational fact. Using a Pari/GP program our series $(1)$ is about $\approx 6.26851$.

$\sum_{k=1}^\infty\frac{\operatorname{rad}(k!)}{\operatorname{lcm}(1,2,\ldots,k)}
$
From its definition,
$rad(k!)
=\prod_{p \le k} p
$
so
$\ln rad(k!)
=\sum_{p \le k} \ln p
=\theta(k)
$
and
$\ln(\operatorname{lcm}(1,2,\ldots,k))
=\psi(k)
$,
where
$\theta$ and $\psi$
are the two
Chebychev 
prime counting functions
(https://en.wikipedia.org/wiki/Chebyshev_function).
Since
$\psi(x)
=\sum_{n=1}^{\log_2x}\theta(x^{1/n})
$
so
$\psi(x)-\theta(x)
=\sum_{n=2}^{\log_2x}\theta(x^{1/n})
\gt\theta(x^{1/2})
\sim x^{1/2}
$.
Therefore
$\frac{\operatorname{rad}(k!)}{\operatorname{lcm}(1,2,\ldots,k)}
=\exp(\theta(k)-\psi(k))
\lt \exp(-k^{1/2})
$
and this sum converges
since,
for any $a>0$,
$k^{1/2}
\gt a\ln(k)
$
for large enough $k$.
Choosing $a=2$
gives
$\exp(-k^{1/2})
\lt 1/k^2
$.
Therefore
the sum converges.
