1. ## Inequalities

Prove that $\frac{\sigma{(n)}}{n} < \frac{n}{\phi(n)} < \frac{\pi^{2}}{6} \frac{\sigma(n)}{n}$ if $n \geq 2$

2. For each prime $p$ let $\alpha_p(n)$ be the greatest $k\in \mathbb{N}$ such that $p^k|n$.

Then $\sigma(n)=\prod_{p|n}\left(1+p+...+p^{\alpha_p(n)} \right) = n\cdot \prod_{p|n}\left(1+p^{-1}+...+p^{-\alpha_p(n)}\right)$ $=n\cdot \prod_{p|n}\left(\frac{1-p^{-(1+\alpha_p(n))}}{1-p^{-1}}\right)$

But remember that $\phi(n)=n\cdot \prod_{p|n}\left(1-p^{-1}\right)$ thus: $\sigma(n)\cdot \phi(n) = n^2\cdot \prod_{p|n}\left(1-p^{-(1+\alpha_p(n))}\right)$ .

Now: $\tfrac{6}{\pi^2} = \prod_{p \text{ prime}}\left(1-p^{-2}\right) < \prod_{p|n}\left(1-p^{-(1+\alpha_p(n))}\right) = \tfrac{\sigma(n)\cdot \phi(n)}{n^2} \leq 1$ - equality holds on the RHS iff n = 1 -

The right inequality is simple enough, each factor is less than 1 - and there are no factors only if n = 1-, now for the left inequality $1-p^{-2} \leq 1-p^{-(1+\alpha_p(n))}$ for each $p|n$, and $1-p^{-2} < 1$ when $p\not | n$ -since n can't have all the primes as divisors, it follows that some inequality must be strict.