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*](https://en.wikipedia.org/wiki/Dedekind_psi_function). On the other hand I add the reference that Wikipedia has the article [*Fermat number*](https://en.wikipedia.org/wiki/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:
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[1] Michal Krizek, Florian Luca, and Lawrence Somer, *17 Lectures on Fermat Numbers*, CMS Books in Mathematics, Springer (2001).



On the behaviour for the quotient involving Fermat numbers of $\frac{\psi(F_m)}{F_m}$ where $\psi(x)$ denotes the Dedekind psi function
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 $1$.


