# 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.

• Many thanks for your nice proof, and congratulations for the credit.
– user243301
Commented Jun 1, 2018 at 8:19