TITLE: How to compute $\text{liminf}\frac{ \varphi(2^n-1)}{2^n}$
QUESTION [5 upvotes]: Let $\varphi$ denote Euler's function. How can I compute $\liminf_n \frac{\varphi(2^n-1)}{2^n}$?

REPLY [2 votes]: If $p_1,...,p_k$ are some distinct prime divisors of $n$, then $\frac{\varphi(n)}{n}\leq \frac{(p_1-1)(p_2-1)...(p_k-1)}{p_1p_2...p_k}$. For the proof see Euler's totient function 
$\forall \varepsilon \in \mathbb R >0 \quad \exists k\in \mathbb N \quad \text{s.t}\quad \frac{(p_1-1)(p_2-1)...(p_k-1)}{p_1p_2...p_k}<\epsilon$ when $p_i$ is the $i$th prime number. Because $\infty=\sum_{i=1}^{\infty}\frac{1}{n}=\prod_p \frac{1}{1-p^{-1}}=\frac{1}{\frac{(p_1-1)(p_2-1)...}{p_1p_2...}}$
You can see Divergence of the sum of the reciprocals of the primes 
If $a|b$ then $2^a-1|2^b-1$
By Fermat's little theorem $p_i|2^{p_i-1}-1$ for all $1\leq i \leq k$   

If we set $n=(p_1-1)(p_2-1)...(p_k-1)$ then by 3 and 4 we can conclude $p_i|2^n-1$
Thus by 1 $$ \frac{\varphi(2^n-1)}{2^n-1}<\varepsilon$$  
and by 2 $$\liminf_n \frac{ \varphi(2^n-1)}{2^n}=0$$