Let $\varphi$ denote [Euler's totient function](https://en.wikipedia.org/wiki/Euler's_totient_function). It's [widely believed](https://en.wikipedia.org/wiki/Mersenne_prime) that $2^n-1$ is prime for infinitely many (prime) $n$, which, in turn, implies
$$\limsup\_{n \to \infty} \frac{\varphi(2^n-1)}{2^n-1} = 1.
$$
But what about an unconditional proof? This is sort of a follow-up of a [question](https://mathoverflow.net/questions/283588/) asked by @kodlu yesterday. 



The limit superior of $\varphi(2^n-1)/(2^n-1)$
Let $n$ be a large prime. Then $2^n-1=p\_1\dots p\_k$ (possibly with equal factors), where $n\mid p\_i-1$ for all $i$. Therefore, $k\leq \log\_n(2^n-1)<\frac{n}{\log\_2n}$. Thus,
$$
 \frac{\varphi(2^n-1)}{2^n-1}\geq\prod\_{i=1}^k\left(1-\frac1{p\_i}\right)
 \geq \left(1-\frac1n\right)^{n/\log\_2n}\to1, \quad n\to\infty,
$$
as required.


