On the behaviour for the quotient involving Fermat numbers of $\frac{\psi(F_m)}{F_m}$ where $\psi(x)$ denotes the Dedekind psi function

Question

In this post we denote the Dedekind psi function as $\psi(m)$ for integers $m\geq 1$. This is an important arithmetic fuction in several subjects of mathematics. As reference I add the Wikipedia Dedekind psi function. On the other hand I add the reference that Wikipedia has the article Fermat number, $F_l=2^{2^l}+1$ and that I was inspired in the results showed in page 101 of [1].

The Dedekind psi function can be represented for a positive integer $m>1$ as $$\psi(m)=m\prod_{\substack{p\mid m\\p\text{ prime}}}\left(1+\frac{1}{p}\right)$$ with the definition $\psi(1)=1$.

Question. I would like (what work can be done about it) if one can to deduce some claim about the behaviour of $$\frac{\psi(F_m)}{F_m}$$ as $m\to \infty.$ Many thanks.

If the question is in the literature please answer the question as a reference request and I try to search and read the results from the literature. If the question is very difficult I ask about the behaviour or heuristic for the quotient $\frac{\psi(F_m)}{F_m}$ for very large integers $m\geq 1$.

References:

[1] Michal Krizek, Florian Luca, and Lawrence Somer, 17 Lectures on Fermat Numbers, CMS Books in Mathematics, Springer (2001).

Answer 1

It is standard that all prime factors of Fermat number $F_m$ are of the form $2^{m+2}k+1$, in particular they are all at least $2^m$. It is then also clear that $F_m$ can only be divisible by at most $2^m/m$ such primes. Therefore $$\frac{\psi(F_m)}{F_m}=\prod_{p\mid m}\left(1+\frac{1}{p}\right)<\left(1+\frac{1}{2^m}\right)^{2^m/m}=\left(\left(1+\frac{1}{2^m}\right)^{2^m}\right)^{1/m}<e^{1/m}.$$ Certainly tighter estimates are possible, but this alone shows the limit of this quantity is equal to $1$.