On a ratio between a number and it's totient I've been having a bit of a problem with a certain ratio:  $ n / \phi(n) $ . While I understand the concept of toitient and the # of relatively prime numbers to n, I have a hard time in figuring out a large number for this ratio to come off of as. This is a bit of a two pronged question with 1. What number could increase this ratio to the point where its more than 10 and how would one find it? (I've tried looking at primes) 2. How would one prove that this number reaches positive infinity?  
 A: We can choose a sequence of integers $n(k)$ such that $\lim\limits_{k\to \infty}\frac{n(k)}{\varphi(n(k))} = \infty$ by letting $n(k)$ be the product of the first $k$ prime numbers.  If $p_1 < p_2 < \cdots$ are all the prime numbers, then $n(k) = p_1 \cdots p_k$, and
$$ \frac{n(k)}{\varphi(n(k))}=\prod\limits_{i=1}^k \frac{1}{1-\frac{1}{p_i}}$$ You can use the geometric series formula for $\frac{1}{1- \frac{1}{p_i}}$ and the fundamental theorem of arithmetic to show that
$$\lim\limits_{k \to \infty} \prod\limits_{i=1}^k \frac{1}{1- \frac{1}{p_i}} = \lim\limits_{m \to \infty} \sum\limits_{i=1}^m \frac{1}{i} = \infty$$
A: If $0 \ne x \in (-1,1)$, then the limit 
$$\lim\limits_{n \to \infty}\sum\limits_{j=0}^n x^j $$
exists, and is equal to $\frac{1}{1-x}$.  Then $$\prod\limits_{i=1}^k \frac{1}{1 - p_i^{-1}} = \prod\limits_{i=1}^k (\lim\limits_{n\to \infty} \sum\limits_{j=0}^{n} p_i^j) = \lim\limits_{n \to \infty} \prod\limits_{i=1}^k \sum\limits_{j=0}^{n} p_i^j$$
where we have used the fact that the limit can be interchanged with a finite product of sequences.  Now,
$$\prod\limits_{i=1}^k(1 + p_i^{-1} + p_i^{-2} + \cdots + p_i^{-n}) = \sum\limits_{(m_1, ... , m_k)} \frac{1}{p_1^{m_1} \cdots p_k^{m_k}}$$
where the sum runs through all $k$-tuples $(m_1, ... , m_k)$ such that $0 \leq m_i \leq n$.  Letting $n$ go to infinity, we see that the product $\prod\limits_{i=1}^k \frac{1}{1-\frac{1}{p_i}}$ is the convergent sum $\sum\limits_{(m_1, ... , m_k)} \frac{1}{p_1^{m_1} \cdots p_k^{m_k}}$, where the $m_i$ are arbitrary nonnegative integers.  
The terms $\frac{1}{d} : d \in \mathbb{N}$ are in bijection with the terms $\frac{1}{p_1^{m_1} \cdots p_k^{m_k}}$, where $k \geq 1$ and $m_1, ... , m_k \geq 0$; this is the fundamental theorem of arithmetic. Then the limit as $k$ goes to infinity of $\prod\limits_{i=1}^k \frac{1}{1-\frac{1}{p_i}}$ is the same as the limit $m$ goes to infinity of $\sum\limits_{d=1}^{m} \frac{1}{d}$.
