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.
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 $p_i^{a_i}.$