P. Erdős and Leon Alaoglu proved in [**1**] that for every $\epsilon > 0$ the inequality $\phi(\sigma(n)) < \epsilon \cdot n$ holds for every $n \in \mathbb{N}$, except for a set of density $0$.


C. L. mentioned in [**2**] that as a consequence of the previous result one can ascertain that $\displaystyle \lim\_{n \to \infty} \frac{\phi(\sigma(n)) }{n} = 0.$


Does anybody know how it is that C. L. proceeded in order to arrive at such a conclusion? 


Clearly enough, the fact that an inequality of the type $a\_{n} < \epsilon \cdot n$ holds for every $\epsilon > 0$ and a subset of $\mathbb{N}$ of density $1$ does not imply, in general, that the sequence $\displaystyle \frac{a\_{n}}{n}$ goes to $0$ as $n \to \infty$.


Hope you guys can shed some light on this inquiry of mine. Let me thank you in advance for your continued support.


**References**


[**1**] L. Alaoglu, P. Erdős: *A conjecture in elementary number theory*, Bull. Amer. Math. Soc. 50 (1944), 881-882.


[**2**] *Mathematical Reflections*, Solutions Dept, Issue #3, 2009, page 23.



A limit involving the totient function
Everyone knows (but no one can prove) that there are infinitely many primes $p$ such that $q=2p-1$ is also prime. $\sigma(q)=q+1=2p$, $\phi(\sigma(q))=\phi(2p)=p-1$, $\phi(\sigma(q))/q=(p-1)/(2p-1)\to1/2$ as 
$q\to\infty$. 


**Edit**: I don't know why it didn't occur to me to look at Guy, Unsolved Problems In Number Theory. Under B42, he writes, "Makowski and Schinzel prove that $\limsup\phi(\sigma(n))/n=\infty$. The reference is A. Makowski, A. Schinzel, On the functions $\phi(n)$ and $\sigma(n)$, Colloq Math 13 (1964-65) 95-99, MR 30 #3870. I haven't found the paper on the web, but it's in Volume 2 of Schinzel's Selecta, 890-894. I don't have the energy to write out the proof in full, but here's the idea. Given $M$, choose $t$ such that 
$$\prod\_{i=1}^t{p\_i\over p\_i-1}>M$$. Then given $p$ (and it's not clear to me whether $p$ is meant to be a prime), and letting $$n=\sigma\left(\prod\_{i=1}^tp\_i^{p-1}\right),$$ we get $$\sigma(n)=\prod\_{i=1}^tN(p\_i,p),$$ where $N(a,p)=(a^p-1)/(a-1)$. Now you prove $\limsup\_{p\to\infty}\phi(\sigma(n))/n\gt M$, using along the way a lemma which says that $$\lim\_{p\to\infty}{\phi(N(a,p))\over N(a,p)}=1.$$


