1. ## Limit Proof

Show that $a_n > 0 ,\ \forall n \in \mathbb{N}$ that $\lim_{n \rightarrow \infty} a_n = 0$ if and only if $\lim_{n \rightarrow \infty} \frac{1}{a_n} = \infty$.

Attempt:
$\lim_{n \rightarrow \infty} a_n = 0 \Rightarrow \lim_{n \rightarrow \infty} \frac{1}{a_n} = \infty$.

so then we have $|a_n - 0| < \epsilon, \ \forall \epsilon > 0$ which would simply imply $|a_n| < \epsilon, \ \forall \epsilon > 0$

If I use the definition:

Let $\alpha$ be in $\mathbb{R}$. The open ray $(\alpha, \ \infty) =\{x \ \in \ \mathbb{R}: \ x>\alpha\}$

so I would get something along the lines of:
$\frac{1}{|a_n|} > \frac{1}{\epsilon}$ then I would have

$\alpha < \frac{1}{|a_n|}$ we can see that $\frac{1}{|a_n|}$ tends to $\infty$

$\lim_{n \rightarrow \infty} \frac{1}{a_n} = \infty \Rightarrow \lim_{n \rightarrow \infty} a_n = 0$.

$\left| \frac{1}{a_n} -\alpha \right| < \epsilon \ \ \ \ \ \ \forall \ \epsilon >0$ this would imply:

$-\epsilon < \frac{1}{a_n} -\alpha < \epsilon \ \ \ \ \ \ \forall \ \epsilon >0$

$(\alpha-\epsilon) \cdot a_n < 1 < (\alpha+\epsilon) \cdot a_n \ \ \ \ \ \ \forall \ \epsilon >0$

at which point I get stuck

2. Here is a very simple outline.
If $\left( {a_n } \right) \to 0$ then $\left( {\forall K \in \mathbb{Z}^ + } \right)\left( {\exists n_K } \right)\left[ {0 < a_{n_K } < \frac{1}{K}} \right]\; \Rightarrow \;K < \frac{1}{{a_{n_K } }}$.

If $\left( {\frac{1}{{a_n }}} \right) \to \infty$ then $\left( {\forall K \in \mathbb{Z}^ + } \right)\left( {\exists n_K } \right)\left[ {K < \frac{1}{{a_{n_K } }}} \right]\; \Rightarrow \;a_{n_K } < \frac{1}{K}$.