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

Question

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.

Answer 1

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 $p_i^{a_i}.$

Answer 2

Is known the infinity production for the Riemann zeta function in the form of $$\zeta(s)=\prod\limits_{\large \forall p\in\mathbb P}\dfrac1{1-p^{-s}}. \qquad(\Re s>1)\tag1$$ Then $$1>\Pi=\prod\limits_{\large p_i}\left(1-p_i^{\large-a_i+1}\right) \ge\prod\limits_{\large p_i}\left(1-p_i^{-2}\right)>\dfrac1{\zeta(2)}=\dfrac6{\pi^2},\qquad(n\ge2).\tag2$$ Using the nice result of Derek Jennings in the form of $$\varphi(n)\,\sigma(n)=n^2\Pi,\tag{$\diamondsuit$}$$ easily to get for $\;n\ge 2:$ $$\color{green}{\mathbf{\dfrac{\sigma(n)}n<\dfrac n{\varphi(n)}<\dfrac{\pi^2}6\,\dfrac{\sigma(n)}n}}.$$