Show using the definition of convergence that the sequence 1/n does not converge to any number $L > 0$. Let $\epsilon > 0$ 
$|1/n - L| < \epsilon$ for some $n\geq N$. (definition of convergence) 
This implies $-\epsilon\cdot n < 1 - nL < \epsilon\cdot n$
Therefore $1 - nL > -\epsilon\cdot n$
Therefore $1 < n(\epsilon - L)$
Choose $\epsilon = 1/n$
Therefore $nL < 2$
Therefore $L < 2/n$ for all $n\geq N$
By Archimedean property, we know that this only holds when $L = 0$. 
Hence the sequence converges to $0$, by uniqueness of convergence the sequence does not converge to any $L > 0$. 
My issue with this proof is that when I choose $\epsilon = 1/n$, technically $\epsilon$ is changing for every choice of $N$. Is this still a valid proof? 
 A: Since you are attempting to prove, by the definition, that the sequence does not converge (to $L$, when $L\neq0$), you are supposed to fix a $\varepsilon>0$. But $\varepsilon$ should be a fixed number neverthelsss.
Take $\varepsilon=\frac{\lvert L\rvert}2$ and let $N\in\mathbb N$. You want to prove that there's a $n\geqslant N$ such that $\left\lvert\frac1n-L\right\rvert\geqslant\frac{|L|}2$. Just take $n$ large enough so that $n\geqslant N$ and also that $\frac1n<\frac{|L|}2$. Then\begin{align}\left|\frac1n-L\right|&\geqslant\left|\frac1n-|L|\right|\\&=|L|-\frac1n\\&>|L|-\frac{|L|}2\\&=\frac{|L|}2.\end{align}
A: $a_n=1/n>0$, decreasing, bounded below by $0$, hence convergent.
Since $a_n >0$, $\lim_{n \rightarrow \infty} a_n=L \ge 0$.
Assume $L>0$.
Let $\epsilon >0$.
There is a $n_0$, positive integer, s.t. for $n \ge n_0$
$|1/n-L| \lt \epsilon$, or
$-\epsilon < 1/n -L < \epsilon$ 
$-\epsilon +L < 1/n$.
Choose $\epsilon =L/2$ (for example),
then $L/2 <1/n$.
Archimedean principle:
There is a $n_1$, positive integer, s.t.
$n_1>2/L$.
For $n > m=: \max(n_0, n_1)$:
$L/2 > 1/m \ge 1/n$,
contradiction. 
Hence $L=0$.
