# Proving $\frac{\sigma(n)}{n} < \frac{n}{\varphi(n)} < \frac{\pi^{2}}{6} \frac{\sigma(n)}{n}$

This is an exercise from Apostol's number theory book. How does, one prove that $$\frac{\sigma(n)}{n} < \frac{n}{\varphi(n)} < \frac{\pi^{2}}{6} \frac{\sigma(n)}{n} \quad \text{if} \ n \geq 2$$

I thought of using the formula $$\frac{\varphi(n)}{n} = \prod\limits_{p \mid n} \Bigl(1 - \frac{1}{p}\Bigr)$$ but couldn't get anything further.

Notations:

• $\sigma(n)$ stands for the sum of divisors

• $\varphi(n)$ stands for the Euler's Totient Function.

• $\sigma$ is the sum of divisors, and $\varphi$ is the totient? – J. M. isn't a mathematician Nov 6 '10 at 12:46
• @J.M: Yes you are rihgt – anonymous Nov 6 '10 at 12:52
• $\frac{\pi^2}{6}$ is $\zeta(2)$ so I wouldn't be surprised if the solution involved comparing integrals or sums. I will work in that direction and come back – marwalix Nov 6 '10 at 13:19
• There is a second part of this problem in Apostol's book, but I can't figure it out. It asks the reader to prove that if $x \geq 2$ then $\Sigma_{n\leq x} \frac{n}{\phi(n)} = O(x)$, where the right hand side of the equation uses the Big-O notation. Any insights? – Jane Roston Nov 22 '11 at 19:19

## 1 Answer

Writing $n=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$ for primes $p_i$ we have

$$\sigma (n) = \prod_{i=1}^k \frac{p_i^{a_i+1} – 1}{p_i – 1}$$

and so

$$\sigma(n)\phi(n) = n^2 \prod_{i=1}^k \left( 1 - \frac{1}{p_i^{a_i+1}} \right) \qquad (1)$$

from which both inequalities follow.

For the RH inequality we note that the expansion of the reciprocal of

$$\prod_{i=1}^k \left( 1 - \frac{1}{p_i^{a_i+1}} \right)$$

is $< \sum_{r=1}^\infty 1/r^ 2 = \pi^2/6.$

Note that

$$\sigma(n) = \prod_{i=1}^k (1+p_i +p_i^2 + \cdots + p_ i^{a_i}),$$

which is where the formula for $\sigma(n)$ comes from and to obtain $(1)$ we've just multiplied this by

$$\phi(n) = n \prod_{i=1}^k \left( 1 - \frac{1}{p_i} \right),$$

and factored out all the $p_i^{a_i}.$