$a_n := 1/\sqrt{n} $. Show that the sequence is bounded, monotone decreasing... I'm given the following:
$(a_n)_{n\in\mathbb{N}} $ is a real sequence defined as $a_n := 1/\sqrt{n} $ for $n \in \mathbb{N}$.
i) Show that $(a_n)_{n\in\mathbb{N}} $ is bounded from above and below, and give the explicit boundaries. 
ii) Show that, $(a_n)_{n\in\mathbb{N}} $, is monotonely decreasing.
iii) Give $\lim_{n\to\infty} a_n  $
Now I understand all of this, Im just not sure how to write a good mathematical proof for it.
Basically my idea is like this... Since n can only be a positive integer, when n = 1, it will basically be  $1/\sqrt{1} = 1$. When n > 1, then the fraction will become smaller, and thats why the above boundary is 1. As n approaches inifinity, then the fraction $1/\sqrt{n}$ will approach 0. So the lower boundary is 0.
The function is monotone decreasing since ($1/\sqrt{n}) \geq (1/\sqrt{n+1})$. 
And $\lim_{n\to\infty} a_n = 0 $ 
This is all clear to me, but I just feel like if I write it like this, it is not really a mathematical proof and I am not sure how I can formulate this as a proper valid proof. 
 A: I think you may want to clean up your answer to (i) a bit, though it is already correct. Here's one possible way to write it:
(i) Let $n \geq 1$. It follows that $a_n$ is trivially positive, and hence $0$ is a lower bound of the sequence $(a_n)_n$. Moreover, $\sqrt{n} \geq 1$ and hence $a_n \leq 1$. As such, $1$ is an upper bound of the sequence $(a_n)_n$.
You should expand upon your answer to (iii):
Hint: (iii) Since the sequence $(a_n)_n$ is nonincreasing with a lower bound of $0$, it must admit a limit. Call this limit $a$. Taking limits on both sides of the inequality $a_n \geq 0$, we get that $a \geq 0$. Now, suppose $a > 0$, and arrive at a contradiction. Conclude that $a=0$.
A: For all $n \in \mathbb{N}\setminus \{ 0\},$ one has $n \geq 1,$ hence $\sqrt{n}\geq \sqrt{1}=1$ and thus $\displaystyle 1 \geq \frac{1}{\sqrt{n}},$ so $a_n$ is bounded above by $1$. We also have (In particular) $n > 0$ for all $n$, hence $\displaystyle \frac{1}{n}>0$ and therefore $\displaystyle \frac{1}{\sqrt{n}}>0$, so the sequence is bounded below by $0$.
Furthermore, for all $n \in \mathbb{N}\setminus \{ 0\}$ we have $n+1 > n,$ hence $\sqrt{n+1}>\sqrt{n}$ and thus $\displaystyle \frac{1}{\sqrt{n}}>\frac{1}{\sqrt{n+1}},$ so the sequence is monotone decreasing.
Finally, given $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\displaystyle \epsilon > \frac{1}{N}$ (Archimedean property), so for all $n \geq N^2$ we have
$$ |a_n-0|=\frac{1}{\sqrt{n}}  \leq \frac{1}{N} <\epsilon,$$
so $\lim_{n \to \infty}a_n=0.$
