Asymptotic of $\sum_{k=1}^n \operatorname{rad}(k!)$ and similar deductions

Question

We denote for integers $m>1$ the product of the distinct prime numbers dividing $m$ as $$\operatorname{rad}(m)=\prod_{\substack{p\mid m\\p\text{ prime}}}p,$$ with the definition $\operatorname{rad}(1)=1$ (see it you want the Wikipedia Radical of an integer).

Definition. We consider to study te following sequence $$\sum_{k=1}^n \operatorname{rad}(k!)\tag{1}$$ for integers $n\geq 1$. The first few terms are $1,3,9,15,45,75,285,495,\ldots$

I believe that it is possible to deduce some staemente about the asymptotic behaviour of this sequence, since the distribution of prime factors of the factorial is well-known.

Question 1. Deduce a statement about the asymptotics of the sequence $$\sum_{k=1}^n \operatorname{rad}(k!)$$ as $n\to\infty$. Many thanks.

A similar question should be to deduce the asymptotic behaviour of sequence of the type $$\sum_{k=1}^n \operatorname{rad}(f(k))\tag{2}$$ where $f(x)$ is a continuous, positive and increasing function on the interval $[1,\infty)$.

Question 2. Is it possible to get a similar statement about the asymptotics of sequences of the type $$\sum_{k=1}^n \operatorname{rad}(f(k))$$ as $n\to\infty$ for examples of continuous and increasing functions $f:[1,\infty)\to[1,\infty)$ that fit the techniques that you've used in deduction of previous Question 1? Many thanks.

Answer 1

We have $\operatorname{rad}(k!)=\prod_{p\leq k}p=k\#$, where $\#$ denotes primorial. We have $k\#=e^{k(1+o(1))}$, and hence we can deduce $\sum_{k=1}^n\operatorname{rad}(k!)=e^{n(1+o(1))}$ as well. You might be worried that this looks like the sum has been approximated by the last term, and that would indeed be correct - any factor which the sum contributes (which is $O(n)$) can be absorbed into the $o(1)$ in the exponent.

Estimates tighter than $o(1)$ in this expression are equivalent to bounding the error term in the prime number theorem, which is famously equivalent to studying the Riemann zeta function (and some estimates, namely replacing $o(1)$ with $O(n^{-1/2+\varepsilon})$, are equivalent to the Riemann hypothesis).

This method is pretty exclusive to $f(k)=k!$, since its prime factors are precisely the primes up to $k$. Any more complicated function (even be it $f(k)=k!+1$) will fail to be subject of this method.